I would like to show that if $n\in\mathbf{N}^{\star} = \mathbf{N}\backslash\{0\}$ is such that $n^2$ divides $2^n + 1$ then $n=1$ or $n=3$. If $n$ is a prime number $p$ it by obvious (and then $p=3$) by writing $2^p = (1+1)^p$ and using the fact that the binomial coefficient ${p}\choose{i}$ is divisible by $p$ when $0<i<p$. I cannot see what happens for a non prime $n$. Knowing that $n$ also divides $2^n + 1$ shows that $n$ belongs to a well-known sequence :
but this doesn't seem to help so far.
Remark. For those for which the question about noting $\mathbf{N}$ or $\mathbb{N}$ is easier than the "real" question, see : https://hsm.stackexchange.com/questions/6726/mathbb-versus-mathbf