The maximum length of time to infect the entire board from $8$ cells is $38$ days and the only arrangement achieving this maximum is as follows (where the $X$'s are infected cells):
--X--X--
X-------
--------
-X------
--------
X-------
-X------
---X---X
Our process can be described as starting with a $2\times 2$ square and adding new cells to enlarge it one step at a time, each time increasing the area's perimeter by $4$. In particular, we start with a square:
X-
-X
which will be infected after $1$ day. We then increase the height of the square by two, placing a new infection such that, once the corner (the last cell infected) of the previous image is infected, the new infection will take effect. In particular, we get:
-X
--
X-
-X
Infecting this takes $3$ days. We then increase the height again in a similar manner yielding:
X-
--
-X
--
X-
-X
Taking $3$ more days. Then, we increase "diagonally", incrementing both height and width:
--X
X--
---
-X-
---
X--
-X-
This takes $6$ days. Then we increase diagonally this again, noting that the last cell filled above is the bottom right corner:
--X-
X---
----
-X--
----
X---
-X--
---X
This takes $7$ days. Then, finally, we increase the width, noting that the top right is the last infected piece, to get:
--X--X
X-----
------
-X----
------
X-----
-X----
---X--
which takes $9$ days to be infected. Then, we increase the width again yielding the correct answer and taking an additional $9$ days. All told, the process will take $38$ days. Moreover, having decided on the sequence of increases to width/height/diagonal, we never had any choice in where to place the infections (up to symmetry) as we always had to put the new infection two cells away from the last infected square of the previous arrangement in a specified direction - once we have shown that the general process generating this answer is optimal, this suffices to show uniqueness of the solution.
Act One: Conclusions Are Proven From Intuitive Assumptions
We will start by taking the following lemma (whose more difficult proof comprises Act Two of this post) and deducing that our construction is optimal. We assume we are on an $n\times n$ board, as our solution generalizes naturally.
The best solution is to start with a $2\times 2$ square, and then add infections such that each successive infection is placed to create a rectangle (of perimeter $4$ larger) with all the previous infections.
In particular, we will describe our above solution as the string:
$$[2\times 2]YYDDXX$$
meaning "start with a $2\times 2$ square, increase its height twice ("Y"), increase it diagonally twice ("D"), then increase its width twice ("X"). After three steps, we could describe the position as
$$[2\times 4]YDDXX$$
as we would have a $2\times 4$ rectangle.
We notice that $[a\times b]X$ will become $[(a+2)\times b]$ after at most $b+1$ days, that $[a\times b]Y$ will become $[a\times (b+2)]$ after at most $a+1$ days and $[a\times b]D$ will become $[(a+1)\times (b+1)]$ after at most $\max(a,b)$ days.
Moreover, we may assume that $a\leq b$ at all times using the symmetries of the problem - in particular, we may assume that, if we start from a square rectangle, then we can exchange every instance of $X$ with $Y$ and vice versa, since the problem clearly has that symmetry (so we should never see the pattern $[a\times a]X\ldots$). This lets us assume that, whenever we have a square of infection, our next move will be $Y$ or $D$ - since if it were $X$, we could exchange all further $X$'s and $Y$'s and get an equivalent length of time. Thus, we may assume $[a\times b]D$ takes $b$ (rather than $\max(a,b)$) days to become $[(a+1)\times (b+1)]$.
Then, we simply notice a few facts to reduce this to a one-variable optimization problem rather than optimization over a space of strings:
$[a\times b]XY$ and $[a\times b]YX$ take equally much time, so we can freely swap their order; without loss of generality, let us assume that the pattern $XY$ never appears (and is always replaced by $YX$).
$[a\times b]XD$ is always slower than $[a\times b]DX$.
$[a\times b]DY$ is always slower than $[a\times b]YD$.
Using the above, we may conclude that the slowest string of enlargements must have the form*:
$$[2\times 2]\underbrace{YY\ldots YY}_{y\text{ times}}\underbrace{DD\ldots DD}_{d\text{ times}}\underbrace{XX\ldots XX}_{x\text{ times}}$$
where we must satisfy $2y+d=n$ and $2x+d=n$ - implying $x=y$ and $d=n-2y$, which reduces this to a one variable optimization problem. In particular, each of the beginning $Y$'s will take $3$ days for the enlargement. The first $D$ will take $2y+2$ days and the last will take $n-1$ days, so they will in total take
$$\sum_{i=2y+2}^{n-1}i=-2y^2 - 3y - 1 +\frac{n^2}2 - \frac{n}2.$$
Then, each $X$ will take $n+1$ days, so they will in total take $y(n+1)$ days. Putting this all together (plus one day for the $2\times 2$ pattern) yields that the length of time taken by the above pattern with $y$ instances of $Y$ will be:
$$-2y^2 + (n+1)y + \frac{n^2}2 - \frac{n}2$$
which is convex, has a maximum at $\frac{n+1}4$ and is symmetrical about that point. Thus, the nearest integer to $\frac{n+1}4$ maximizes that. So, for $n=8$, we get that $y=2$ (rounded from $\frac{9}4$) is optimal, and the above expression yields $38$ as the upper bound at that point. In particular, we can find the optimal solutions for all $n\geq 4$ as:
If $n$ is between $4c-3$ and $4c+1$, then an optimal solution is to enlarge upwards $c$ times, then diagonally $n-2c$ times, and then rightwards $c$ times. (If $n$ is not $1$ mod $4$, this $c$ is unique).
Similar methods apply to a $n\times m$ square with $n\leq m$, where $n+m$ is even and we have $\frac{n+m}2$ infections).
Act Two: We Replace Our Intuition with Mathematical Contortionism
Unfortunately, it takes a lot more machinery to prove that our lemma was correct than it does to apply it - in particular, as natural as it is to talk about rectangles, this structure is not implicit in the update rule of our automata, yet we need to apply the model as an upper bound nevertheless. Firstly, we need to define what these rectangles are. Our goal is to, for any infection, create a binary tree of rectangles, such that each node is the rectangular infection generated by combining the rectangles of its two children nodes, and then to characterize the length of an infection based on the properties of this tree. We do not directly construct such a tree, but it is implicit in the induction we use, and the major lemma used in this section may be used to construct it.
In particular, let us begin with the following definition:
Let $S$ be a finite subset of $\mathbb Z^2$. Define $\newcommand{\cl}{\operatorname{cl}}\cl(S)$ to be the set of points eventually infected if those in $S$ are infected.
It should be clear that $\cl$ is a closure operator. Moreover, the closed sets with respect to this operation (i.e. sets such that $\cl(S)=S$) are exactly those from which no new infections are possible - we may characterize these as follows:
A set $S$ is closed if and only if it a union of rectangles, no cell being adjacent or contained in more than one rectangle.
Now, to build our tree of infections, we use the following important lemma:
For all $S$ with more than one element such that $\cl(S)$ is a rectangle, there exists a disjoint pair $A,B\subseteq S$ such that $\cl(A)$ and $\cl(B)$ are rectangles neither equal to $\cl(S)$ and that $\cl(A\cup B)=\cl(S)$.
To prove this, consider the set of rectangles other than $\cl(S)$ which are closures of subsets of $S$. This is obviously finite and non-empty if $|S|>1$. Therefore, ordered by inclusion, there is a maximal element $\cl(A)$ - a rectangle not contained by any other such rectangles. Set $B=S\setminus \cl(A)$. Now, suppose that $\cl(B)$ is the union of rectangles $\cl(B_1),\cl(B_2),\ldots,\cl(B_k)$. Notice that $\cl(A\cup B_i)$ must strictly contain $\cl(A)$. It is therefore either equal to $\cl(S)$, in which case we're done, or it is not a rectangle. If none of the $B_i$ have that $\cl(A\cup B_i)$ was a rectangle, then it would be the case that no cell could be adjacent to $\cl(A)$ and a $\cl(B_i)$ (nor to distinct $\cl(B_i)$ and $\cl(B_j)$), implying that $\cl(A)\cup\cl(B_1)\cup \cl(B_2)\cup\ldots\cup \cl(B_k)$ was a closed set containing $S$ which was not a rectangle, contradicting that $\cl(S)$ is a rectangle. Thus, one of the $B_i$ must satisfy the conditions of the lemma. (If $S$ has that its perimeter and it's closure's perimeter are equal, we may easily modify the above proof to show that $A$ and $B$ can and must be a partition of $S$)
Using this tree, we may bound the length of time it takes for the infection to complete its spread. In particular, let $t(n,m)$ be the maximum length of time it takes to infect all the corners of an $n\times m$ rectangle with $\frac{n+m}2$ initial infections, where $n+m$ is even. This equals the amount of time to infect the whole square for $n,m>1$, since all the possible combinations of rectangle infect a corner last and only begin once both sub-squares have their corners infected. Notice that there are only $3$ manners of combining two rectangles such that the result has the sum of their perimeters - on two $2\times 2$ rectangles, we illustrate them as follows:
-XX
-XX
---
XX-
XX-
--XX
--XX
XX--
XX--
---XX
XX-XX
XX---
Which correspond to increasing the height, the diagonal, and the width. In particular, to make an $n\times m$ rectangle ($n,m>1$), one must choose one of five options. For the first $3$, suppose that $x_1,x_2,y_1,y_2$ are positive integers with $x_1+y_1$ and $x_2+y_2$ both even. We are combining rectangles $x_1\times y_2$ and $x_2\times y_2$ in one of the above manners, letting $T=\max(t(x_1,y_1),t(x_2,y_2))$.
Combining in the first manner (heightwise). We require $x_1+x_2=n+1$ and $y_1+y_2=m-1$ and $x_1,x_2>1$. This takes $T+\max(x_1+y_2,x_2+y_1)$ days.
Combining in the second manner (diagonally). We require $x_1+x_2=n$ and $y_1+y_2=m$. This takes $T+\max(x_1+y_2,x_2+y_1)-1$ days.
Combining in the third manner (widthwise). We require $x_1+x_2=n-1$ and $y_1+y_2=m+1$ and $y_1,y_2>1$. This takes $T+max(x_1+y_2,x_2+y_1)$ days.
Increasing height only. $x_1=1$ and $x_1+x_2=n+1$ and $y_1+y_2=m-1$. This takes $T+x_2+y_1$ days.
Increasing width only. $y_1=1$ and $x_1+x_2=n-1$ and $y_1+y_2=m+1$. This takes $T+x_1+y_2$ days.
Which, taking $t(1,n)=t(n,1)=0$, yields a recurrence relation for $t$ allowing explicit computation. However, even better than that, we can show that the first three options are never optimal, hence we only use the last two options - but moreover, those options are obviously slowest when $x_1=y_1=1$, meaning we should always enlarge rectangles according to the previously used procedure.
To prove this, assume without loss of generality that $t(x_1,y_1)\leq t(x_2,y_2)$ - so essentially, the creating of the first rectangle comes for free, since it was done in parallel with the second rectangle. For the first (and by symmetry, the third) case, suppose we instead of joining a $x_1\times y_1$ rectangle to the $x_2\times y_2$ one, we increase the height of the $x_2\times y_2$ rectangle with a $1\times 1$ square placed above it, and then diagonally join the resulting $x_2\times (y_2+2)$ rectangle to a $(x_1-1)\times (y_1-1)$ rectangle**. That is, we consider the alternate diagram:
--X
-X-
---
XX-
XX-
This takes a total of $T+[x_2+1]+[\max(x_1-1+y_2+2,x_2-1+y_1)-1]$
which is at least as long as before. Similarly, when combining rectangles diagonally, we instead join a $1\times 1$ rectangle to the $x_2\times y_2$ one and then join the $(x_1-1)\times (y_1-1)$ one to that. We find a similar result algebraically.
The above suffices to show the desired claim that the optimal solution is of the form claimed, implying the displayed solution is optimal and unique for a $8\times 8$ board.
(*If we we're being very careful, we would note that all of the exchanges suggested, including exchanging all $X$'s and $Y$'s after something of the form $[a\times a]X\ldots$, decrease the lexicographic order of the string when we take $Y<D<X$. Thus, we will eventually reach a minimal element if we greedily apply such swaps - and in this context, a minimal element is one in which no such swaps are possible and is of the given form)
(**Some special care needs to go into the $y_1=1$ case, but the result comes out the same. A similar exception presents itself so for the diagonal combination)
