Determine all integers $n>1$ such that $\frac{2^n+1}{n^2}$ is an integer.
Attempt:
Notice that $n$ must be odd. And $n=3$ satisfies the condition. The number that we want must therefore satisfy
$$ 2^{n} \equiv (-1) \bmod (n^{2})$$
or
$$ 2^{n} = \alpha n^{2}-1 = (\sqrt{\alpha} n -1)(\sqrt{\alpha}n+1).$$
Notice that $\sqrt{\alpha}$ must be odd.