(This proof was completed with an insightful comment from Gottfried. I'm placing it as an answer so that it is readily seen that an answer exists, as opposed to just leaving it in the question.)
Suppose we have $ 2^n \pm 1 = x^k$ for some positive integers $x, k$. Cases $n=1, 2, 3$ are easily dealt with. $n=3$ yields the solution $2^3 + 1 = 3^2$. Henceforth, let $n\geq4$. Furthermore, a simple check shows that $x\neq 1, 2$.
First, we will deal with the power $k=2l$ being even. Rewriting $x^{2l}$ as $(x^l)^2$, we may assume that $k=2$.
Proof: $2^n -1 \equiv 3 \pmod{4}$ hence is never a square.
If $2^n +1 =x^2$, then $2^n = (x-1)(x+1)$ and both of these are powers of 2 that differ by 2. Thus we must have $(x-1) = 2, (x+1) = 4$, which gives $2^n = 8$ so $n=3$ (reject). $_\square$
Now, we deal with $k$ odd.
Proof: Say $2^n +1 = x^k$. Then $2^n = x^k - 1 = (x-1)(x^{k-1}+x^{k-2} + \ldots +1)$, so both terms are powers of 2. We have $ x -1$ is a power of 2 greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $k$ odd numbers, hence is odd. Since this is clearly greater than 1, we get a contradiction.
Say $2^n -1 = x^k$. Then $2^n = x^k + 1 = (x+1)(x^{k-1} - x^{k-2} + \ldots -x +1 )$, so both terms are powers of 2. We have $x+1$ is a power of 2 that is greater than 1, hence is even. Thus $x$ is odd. But the other term is the sum of $p$ odd numbers, hence is odd. Since $x^p + 1 \neq x+1$ except for $p=1$, this means that the term is greater than 1. Hence we get a contradiction. $ _\square$
The total number of handshakes overall has to be even.
Suppose the number of people who have shaken an odd number of hands is odd. Odd times odd is odd. So the total handshakes in this group is odd.
The number of people who have shaken an even number of hands can odd or even. But the total hands shaken by them will always be even times even or even times odd = even.
Adding them up gives a contradiction that says "odd + even = even".
Edit: You need to consider the sum of hands shaken in both groups instead of just the "odd handshake group" because the "odd group" could have shaken hands with people outside of their group too and so is allowed to have shaken an odd number of hands.
i.e. It is not true that the total number of hands shaken by a subset of the group has to be even.
Best Answer
You've simply stopped calculating one step too soon! You've shown that that expression, with $32!$ divided out is divisible by $2$, but if you just checked whether it was divisible by $4$, you would see that it is indeed not. In particular, the divided expression would be $$\frac{32!}{32!}+\frac{33!}{32!}+\frac{34!}{32!}+\frac{35!}{32!}+\frac{36!}{32!}+\ldots+\frac{90!}{32!}$$ Now, take this mod $4$. All but the first four terms are eliminated, since $4$ divides each of them (as each has $36$ as a factor). However, the first four terms, mod $4$ are $1,\,1,\,2,\,2$, which sums to $2$ mod $4$. Ergo, $2^{33}$ does not divide the original expression, and $2^{32}$ is thus the maximum power of two dividing the expression.