What you write is not correct: the number of elements of order a power of $2$ is not necessarily equal to $2^5-1$ times the number of $2$-Sylow subgroups because, as Zev Chonoles points out, you do not have that two distinct $2$-Sylow subgroups must intersect trivially (which is what goes behind that particualr inequality).
The number of $31$-Sylow subgroups must divide $31\times 32$ and be congruent to $1$ modulo $31$; so either there is a single $31$-Sylow subgroup (in which case the group is not simple), or there are thirty two $31$-Sylow subgroups.
If there are thirty two $31$-Sylow subgroups, then since any two distinct ones must intersect trivially (the groups are cyclic of prime order, so the only proper subgroup is trivial), they account for $32(31-1) + 1$ elements of $G$.
That means that there are $32\times 31 - 32\times 30 = 32$ elements whose order is not $31$. Since a $2$-Sylow subgroup must contain $32$ elements, there are only enough elements left over for a single $2$-Sylow subgroup, which must therefore be normal.
So $G$ will have either a single $31$-Sylow subgroup, or a single $2$-Sylow subgroup. Either way, it is not simple.