( Jaded by mathematics?
At the end of this answer : 
The Case of the Surprisingly Simple Solution )
Bonus task: Find some nice, easy-to-explain encodings.
$ \require{begingroup}\begingroup
\def \Pad #1{\raise{#1}{\small\raise1mu\strut}}
\def \Frac #1#2{ \dfrac {\raise-9mu {#1}} {\raise2mu {#2}} }
\def \SSfrac #1#2{ \Frac { \scriptsize #1 } { \scriptsize #2 } }
\def \Sfrac #1#2{ \Frac { \small #1 } { \small \raise4mu {#2}} }
\def \T #1{{ \small\textsf{#1} }}
\def \Nsub #1{ n \raise-3mu{\kern1mu\T{#1}} }
\def \Tight {\kern -1mu}
\def \Hair {\kern 1mu}
\def \Hairr {\kern 2mu}
\def \MinusL {{\kern 1mu {-} }}
\def \MinusLR {{\kern 1mu {-} \kern 1mu}}
\def \Minus {{\kern 1mu {-} \kern-1mu}}
\def \Minuss {{\kern-1mu {-} \kern-1mu}}
\def \MinusR {{ {-} \kern 1mu}}
\def \MinusRR {{ {-} \kern 2mu}}
\def \MinusXR {{\kern-1mu {-} \kern 1mu}}
\def \PlusL {{\kern 1mu {+} }}
\def \PlusR {{ {+} \kern 1mu}}
\def \Plus {{\kern 1mu {+} \kern-1mu}}
\def \Pluss {{\kern-1mu {+} \kern-1mu}}
\def \EB { \T {equivalent bits} }
\def \AEB { \T {average equivalent bits} }
\def \TEB { \T {total equivalent bits} }
\def \PC { \T{punch cells} }
\def\Ttimes {{ \tiny \kern1mu \raise2mu\times \kern2mu }}
\def \SStimes {{ \scriptsize \raise1mu\times }}
\def \Stimes {{ \small \kern-1mu\raise1mu\times \kern-1mu }}
\def \Logbb { \log_2 }
\def \Logb { \log_2 \kern-2mu }
\def \Arrow #1{\mathop{ \raise-11mu\rlap{ \displaystyle\small~~~#1 }
\raise-1mu\xrightarrow{\hphantom{ \displaystyle\small~~#1 }} }}
$
1.
A degenerate
trivial-to-explain encoding happens to be unbeatable.
Any nonnegative integer $N$ can be conveyed,
and the measure in question increases boundlessly with $N$ itself.
$$ n ~ = ~~ \Frac{\EB}{\PC}
~~ \approx ~~ \Logb N ~~\Arrow{N\to\infty}~~~~ \infty $$
Just one punchable cell can be reused any number of times
to convey an arbitrarily large $N$.
•  
First send the unpunched card $N$ times.
(Each time amounts to a
unary 1.)
•  
Then punch that cell and send the punched card once.
End of number. End of card.
Although this feels like cheating,
similar iteration is required for decreasing-base approaches too,
unless a card is considered usable only when a value to be conveyed
happens to be within the card's one-shot capacity at the moment.
2.
A more interesting and only slightly less easy-to-explain encoding
for fixed-base values seems able, on random average,
to get asymptotically maximal use from a punch card.
As an example,
a single 27-cell card can convey 25 decimal places of $\,\ln 2$,
i.e, 0.6931471805599453094172321,
one digit at a time, with only 25 holes punched.
•  
The card's 27 punch cells are weighted
1, 1, 1, 2, 2, 2, 3, 3, 3, . . . , 8, 8, 8, 9, 9, 9.
•  
To read the current value, take the last digit
of the sum of all currently-punched cells' weights.
•  
To store a new value,
punch the fewest not-yet-punched cells
whose weights add up to the difference,
modulo 10,
between the current and new values.
Except, though, if the new value matches the old value,
when no new holes need to be punched.
And if no combination of new holes can achieve the new value then the
card is considered used up.
$$ n ~ = ~~ \Frac {\TEB}{\PC}
~~ = ~~ \Sfrac{ 25 \Logb 10 }{ 27 }
~~ = ~ 3.1
$$
Not bad, as the most information this card could possibly convey,
even if its history were in play, is a little more than the
equivalent of 31 decimal digits for
$~ \Nsub{perfect} = 3.8 ~$ (elaborated later).
Calling $n$ of the generalized approach here
$\Nsub{additive} \Hair$,
if the formulation that follows is valid then,
for an arbitrary $m$,
a card with $~ c = m \Hair 2^m \,\Pad{2mu}$ punchable cells yields
(“$\small \sim$” being
asymptotic equivalence):
$$ \Nsub{additive} ~~\Arrow{ c = m \Hair 2^m \to \, \infty }~~~ m
~~\sim~~ \Logb c
~~\sim~~ \Nsub{perfect}
$$
This approach always encodes a
base-$ \raise1mu{\small(} 2^m
\raise2mu{{\scriptsize\kern-1mu +}}
1 \raise1mu{\small)}$
value from $0$ through $2^m$ and, for the sake of simpler calculations,
differs in inconsequentially detrimental ways from the 27-cell example above.
•  
Each cell is weighted by a random integer
from $1$ through $2^m \Tight$.
•  
To read a card's current value,
add up the weights of all punched cells, modulo $2^m\Pluss1$.
•  
To store a new value, when different from the current value,
punch one not-yet-punched cell
whose weight is the difference, modulo $2^m\Pluss1 \Pad{-2mu}$.
If that weight is not available for punching,
revise the card's current value by punching a random hole,
then try again to store the same new value.
Here comes a table with supporting formulations,
optimistically without rigor. For convenience:
$\kern2em \llap{b} ~ = ~~ \rlap{ 2^m + 1 } \Pad{6mu}\Pad{-6mu}
\kern7em$ (fixed base of values, which range from $0$ through $2^m$)
$\kern2em \llap{p} ~ = ~~ \rlap{ \Sfrac{1}{b\Minus1} ~~ = ~~ m{/}c }
\kern7em$ (probability that a random non-0 value
matches a given non-0 value)
$\kern2em \llap{q} ~ = ~~ \rlap{ 1 - p } \Pad{4mu}
\kern7em$ (probability that a given cell has an undesired weight)
$$
\def \L {\kern-7mu} \def \R {\kern-16mu}
\def \Star { ~~~~~~ \color{#9c0}{\huge\raise-1mu\star} }
\def \U #1{ \underline{\strut~{#1}~} }
\def \Thirds #1#2#3#4#5{ \kern#1 \llap{#2} {#3} \rlap{#4} \kern#5 }
\def \POf #1#2{ \Thirds {1.6em} {#1} {~\T{of}\,~} {#2} {1.4em} }
\def \PHalves #1#2{ \rlap{ \kern 4.2em \llap{#1~~} \rlap{#2} } }
\def \PUses #1#2#3{ \Thirds {4.2em} {#1} {#2} {#3} {4.1em} }
\def \AOf #1{ \Thirds {1.7em} {#1} {~\T{of}\,~} {b} {1.7em} }
\def \AHalves #1#2{ \rlap{ \kern 5.1em \llap{#1~~} \rlap{#2} } }
\def \AUses #1#2#3{ \Thirds {6.8em} {#1} {#2} {#3} {2.9em} }
\def \POfT #1#2{ \POf{\T{#1}}{#2} }
\def \PUsesTimes #1#2{ \PUses {#1 \kern1mu} {\kern2mu \SStimes \,} {#2} }
\def \PUsesLog #1#2{ \PUsesTimes { \big( 1 \PlusR \Sfrac{1}{#1} \big) \! }{ \Logb #2 } }
\def \PUsesOne #1{ \PUsesTimes { \Sfrac{1}{1} \kern#1 }{ 1 } }
\def \AOfT #1{ \AOf{\T{#1}} }
\def \AUsesTimes #1#2{ \AUses { #1 \big( #2 \big) \kern-1mu } {\kern2mu \SStimes \,} {\kern1mu \Logb b } }
\def \AUsesQ #1{ \AUsesTimes { ( 1 \MinusXR #1 ) } { 1 \PlusR \Sfrac{1}{b\Minus1} } }
\begin{array}{c|cc|cc}
\L & \U{\Nsub{perfect}} & \U{\Nsub{perfect} } & \U{\Nsub{additive}} & \U{\Nsub{additive}} \R\\[-1mu]
\L \T{Number of} & \T{ Number of } & \T{ Expected } & \T{ Number of } & \T{ Expected } \R\\[-5mu]
\L \T{unpunched} & \T{ possible } & \T{number of uses } & \T{ possible } & \T{number of uses } \R\\[-5mu]
\L \T{ cells } & \T{ values } & \SStimes ~ \T{equivalent bits} & \T{ values } & \SStimes ~ \T{equivalent bits} \R\\[11mu]
\hline
\L c & \POf { 1 }{ 2 } & \PUsesOne{ 14mu } & \AOf {1} & \AUsesTimes{\Sfrac{1}{b}~}{\Sfrac{1}{b\Minus1}} \R\\
\L c-1 & \POfT {all}{ c } & \PUsesLog{c\Minus1}{ c } & \AOf {b\Minus1} & \AUsesQ { q^ c } \R\\
\L c-2 & \POfT {all}{c\Minus1} & \PUsesLog{c\Minus2}{(c\Minus1)} & \AOfT {all} & \AUsesQ { q^{c\Minus1} } \R\\[-.5ex]
\L \vdots & & \PUses {}{\vdots}{} & & \AUses {\vdots\kern34mu}{}{} \R\\[.5ex]
\L 2 & \POfT {all}{ 3 } & \PUsesLog{ 2 }{ 3 } & \AOfT {all} & \AUsesQ { q^3 } \R\\
\L 1 & \POfT{both}{ 2 } & \PUsesLog{ 1 }{ 2 } & \AOfT {all} & \AUsesQ { q^2 } \R\\
\L 0 & \POf { 1 }{ 2 } & \PUsesOne{ 6mu } & \AOfT {all} & \AUsesQ { q } \R\\[1ex]
\hline
\L \Pad{1ex}
\L \T{ Total } & \PHalves{ c \Ttimes \Nsub{perfect} ~ = }{ } & & \AHalves{ c \Ttimes \Nsub{additive} ~ = }{ } & \R\\[-.5ex]
\L \raise2ex\T{bits} & & \kern8.9em \llap{ \displaystyle 2 + \sum_{i=2}^c \Logb i + \sum_{i=2}^c\Sfrac{\Logb i}{i \MinusL 1} } & & \AUsesTimes{ \Sfrac{1}{b^2{-}b} + \Big( c - \Sfrac{q \MinusR q^{c \PlusL 1}}{1 \MinusRR q} \Big) }{ \Sfrac{b}{b\Minus1} } \R\\[.5ex]
\L & \PHalves{ \sim }{ \Logbb c \Hairr ! } & & \AHalves{ \Arrow{c = m \Hair 2^m \to \, \infty} }{ \big( c \MinusL \Sfrac{1}{\raise-1mu p} \big) \Hair \Logbb (2^m\Pluss1) } & \R\\[1ex]
\L & \PHalves{ \sim }{ \Logb \Hair \sqrt{2 \pi c} \, \big( \Sfrac{c}{e} \big)^c } & & \AHalves{ \sim }{ c \Hairr m - \Sfrac{m}{m{/}c} } & \R\\[1ex]
\L & \PHalves{ \sim }{ c \Hair \Logb c } & & \AHalves{ \sim }{ c \Hairr m } & \R\\[2ex]
\hline
\L \Pad{1ex}
\L n & \PHalves{ \Nsub{perfect} ~ \sim }{ \Sfrac{c \Hair\Logb c}{c} } & & \AHalves{ \Nsub{additive} ~ \sim }{ \Sfrac{c \Hairr m}{c} ~~ = ~~~ m \Star } & \R\\[-.5ex]
\L & \PHalves{ = }{ \Logb c \Star } & & \AHalves{ = }{ \Logb 2^m } & \R\\[.7ex]
\L & & & \AHalves{ \sim }{ \Logbb (m \Hair 2^m) } & \R\\[.7ex]
\L & & & \AHalves{ = }{ \Logb c ~~~~~~ \Star } & \R\\
\end{array}
$$
Sure enough, looks like
$~ \Nsub{additive} ~\sim~ m ~\sim~ \Logb c ~\sim~ \Nsub{perfect} \Hair$.
About this table
“$\T{Number of possible values}$”
is interestingly anomalous for unpunched cards
as well as for fully punched cards.
Their values here are debatable,
but at least they don't affect asymptotic behavior.
“$\T{Expected number of uses}$”
combines one or two probabilities.
In the row for $2$ unpunched cells, e.g:
i.
$(1 \MinusRR q^3) ~\Pad{11mu}\Pad{-4mu}$
under $\Nsub{additive}$
is the probability that the desired value-difference weight
happened to be available for punching when the card had $3$ unpunched cells.
ii.
$\big( 1 \PlusR \Sfrac{1}{b\Minus1} \big) ~\Pad{16mu}\Pad{-12mu}$
under $\Nsub{additive}$ indicates that,
whenever a card is used as punched,
it gets an average of $~ \Sfrac{1}{b\Minus1} ~$
additional uses for free,
as subsequent values might repeat the current value. The corresponding
$~ \big( 1 \PlusR \Sfrac{1}{2} \big) \Hair\Pad{9mu}\Pad{-9mu}$
under $\Nsub{perfect}$ varies by row,
because so does the number of values that would not be repeats.
By the way,
$~ \Sfrac {1}{b\Minus1}
= \Sfrac {1}{b} \Stimes\Sfrac {b\Minus1}{b}\Stimes\Hair 1 \,\T {extra use}
+ \Sfrac{1}{b^2}\Stimes\Sfrac{b\Minus1}{b} \Stimes\Hairr 2 \,\T{extra uses}
+ \cdots \,$.
About that 27-cell example and $\Nsub{perfect}$
For $~ c = 27 \Pad{21mu}$, the table gives
$~ \displaystyle c \Ttimes \Nsub{perfect}
= 2 + \sum_{i=2}^{27} \Logb i + \sum_{i=2}^{27} \Sfrac{\Logb i}{i \MinusL 1}
= 103.5 $ total equivalent bits,
so $ \Pad{5mu}\Pad{-11mu}
~ \Nsub{perfect} = \Frac {\TEB}{\PC} = \Sfrac {103.5}{27} = 3.8 \Hair $.
Likewise, the formulation of $\, c \Ttimes \Nsub{additive} \,$
predicts an expected average measure,
instead of the $~ n=3.1 ~$ achieved in the example, of
$~ \Nsub{decimal}
= ~ \Big( \Sfrac{1}{90}
+ \Bigg( 27 - \Sfrac{\SSfrac89 \MinusLR \big( \SSfrac89 \big)^{\! 28}}
{\SSfrac19}
\Bigg) \Sfrac{10}{9}\Logb 10
\Big) { \large / } 27
= 2.6^+ \Pad{28mu}\Pad{-24mu}$,
which is ever so close to the
$~ n = \Sfrac{ 21 \Logb 10 }{ 27 } = 2.6^- ~\Pad{-14mu}$
that corresponds to the 21 decimal places conveyed before
the two holes for the 22nd digit were selected nonrandomly.
Side note: The Degenerate Case of the Fourfold Use
Back to the smaller-is-better idea of a 1-cell card,
binary use of a single-cell card works surprisingly well too.
Using 27 single-cell cards for the same value as in the 27-cell example,
$\ln 2 \Hair$, a whopping 121 fraction bits are conveyed:
0.1 011 0001 0111 001 00001 01111111 011111 01 000111 001111 01111 0011 01 01
01111 001 001111 000111 011 00111 0011 000000000111111 001 01111 011 01.
That's equivalent to 36.5 decimal places
and a surreal
$~ n_{\small 27} \, = \Sfrac{121}{27} = \, 4.5 ~$ bits/cell. However:
•  
The segment 000000000111111 is quite an outlier.
Disregarding its 15 bits makes some difference, leaving
$n_{\small 26} \, = \Sfrac{106}{26} = \, 4.1$ bits/cell.
•  
For binary, $~ n \approx 4 ~$ actually does make sense.
Each value can be expected to have an average of one
additional reuse due to consecutively repeated values.
And the hole doesn't need to be punched for the initial value
of any card except the very first,
as that value will be $0$ for each subsequent card.
Epilogue: The Case of the Surprisingly Simple Solution
It all began like so many puzzles here.
No sooner had a poster posed a puzzle with an alluring premise than a
thrilling solution
was pulled off in broad daylight by a f''amiliar perpetrator.
A solution so swift and devious
that the original poser felt compelled to display their
ur  solution
as well.
Not much else left to do, apparently,
except to strew some smarty-pants comments.
With that accomplished, something nonetheless seemed amiss.
Was a different solution still on the loose?
The poser dangled fresh bait by mounting a
companion piece
on a fixed base.
The puzzle became an adventure when a discarded early
hunch —okay, a smarty-pants
comment— returned
as a clue, leading to the hidey-hole of the one-eyed punch card.
(Encoding #1 at the top of this post.)
For reassurance that all solutions had at last been rounded up,
Sherlock Holmes outlined a crudely outlandish scheme as an
impossibility to be eliminated.
But Scotland Yard botched the mathematical labwork
and mistakenly considered this straw-man scenario plausible after all
(a nascent encoding #2 of this post).
Ensuing paperwork inevitably revealed many flaws,
so the investigation was appropriately shelved.
While off duty, the inspectors regrouped,
tried to rework the embarrassingly improbable
into the possibly probabilistic,
and seemed to be onto something,
but didn't know how to present probable cause to the grand jury.
For instance, the detectives had zeroed in on some
prime evidence
on $\large e$ Street but the puzzle had been committed
some distance away, at the $\small\rm Log_2\!$ Club.
In the spirit of peccable puzzle-police work,
this was resolved by disguising that evidence
and planting it at the original scene.
For this, two bit players, $b$ and $c$,
plucked from a number-lineup,
agreed to help in exchange for a sweet taste of
power (2m&m's).
So encoding #2 was assembled from a variety
of surprising pieces, some unmentioned.
Many of the steps were characteristic of a bumbling
Inspector Clouseau
stumbling over serendipitous answers
while trying to complicate things with one
shot in the dark
after another.
After chancing on some optimal pieces,
a puzzle-within-a-puzzle was to find similar-enough pieces
that would fit together without toooo much mathematical mess.
Thanks to the largesse of large numbers
and the ratio-nullifying nature of logarithms,
some intuitively contrived formulations
could stand in for optimal ones that they only vaguely resemble.
(Then again, the later labwork may contain mistakes too.)
$\endgroup$