The general picture here is as follows: if you have two positive integers $m,n$ with no common factor then every integer bigger than $mn-m-n$ can be written as $am+bn$ with $a,b$ non-negative integers, but $mn-m-n$ itself can't.
First, suppose $mn-m-n = am+bn$. Then $(b+1)n$ is a multiple of $m$, and therefore (since $m,n$ have no common factor) so is $b+1$; so $b \geq m-1$. Similarly, $a \geq n-1$. So $am+bn \geq 2mn-m-n$, contradiction.
So $mn-m-n$ isn't $am+bn$ with $a,b$ non-negative integers.
[In this particular case, this says: suppose $63=5a+17b$. Then $80=5a+17(b+1)$, so $17(b+1)$ is a multiple of 5, so $b+1$ is too, so $b\geq4$. And $68=5(a+1)+17b$, so $5(a+1)$ is a multiple of 17, so $a+1$ is too, so $a\geq16$. But then $5a+17b\geq80+68>63$, contradiction.]
Now, to show that any integer $k\geq mn-m-n+1=(m-1)(n-1)$ will do, we'll first do it allowing negative numbers of stamps, and then fix up the negativity.
It turns out that if you allow a negative number of stamps, you can do this for any $k$. Let's do it for $k=1$; that is, find $a,b$ not necessarily non-negative with $am+bn=1$. One way to do it is to run Euclid's algorithm on $m,n$ and keep track of how to write every number involved in the form $am+bn$. At the last step, one of those numbers is 1 and you're done.
[In this particular case, we'll find e.g. that $7\times5-2\times17=1$.]
So now we have $k=(ka)m+(kb)n$, but of course our coefficients may be negative. The remaining job will be to fix it up.
[In this particular case, let's take $k=64$; we get $64=448\times5-128\times17$.]
So, how to fix it up? If $am+bn=k$ then for any integer $t$ we have $(a+tn)m+(b-tm)n=k$. We'll try to choose $t$ to make both $a+tn$ and $b-tm$ non-negative. That means making $t\geq -a/n$ and $t\leq b/m$. If we can't do this then there's no integer between $-a/n$ and $b/m$, which means $b/m-(-a/n)<1$; that is, $b/m+a/n<1$; that is, $am+bn<mn$. The LHS here is our $k$, but (alas!) this isn't quite the inequality we're looking for; we'll have to be more precise.
The trick here is that $a+tn$ and $b-tm$ are always integers. So making them non-negative is the same as making them $>-1$; what we need, then, is $a+tn>-1$ and $b-tm>-1$; equivalently, $t>(-a-1)/n$ and $t<(b+1)/m$. If there is no integer (strictly) between $(-a-1)/n$ and $(b+1)/m$ then that interval must be no longer than 1 unit, so $(b+1)/m+(a+1)/n\leq1$. Multiplying through by $mn$ and using $am+bn=k$ gives us $k<=mn-m-n$, and we're done.
[In this particular case with k=64 as above, what we need is $448+17t\geq0$ and $-128-5t\geq0$, which we tweak to $448+17t>-1$ and $-128-5t>-1$, which are equivalent to $t>-449/17$ and $t<-129/5$. Those bounds are approximately -26.4 and -25.8 respectively, and as you can see -25 lies between them.]
What if the two positive integers $m, n$ do have a factor in common? In the case of e.g. 15 and 21, both denominations are divisible by 3. This in turn means that any combination will be divisible by 3, and thus there are infinitely many values that you cannot make up by a combination. In general, if the greatest common divisor of $m,n$ is $d$, then you can only make multiples of $d$; the largest that you can't will be $d((m/d)(n/d)-(m/d)-(n/d))=mn/d-m-n$.