Solution:

$\frac{2^n+1}{n^2}$ is an integer $\implies n^2 \mid 2^n+1$

Let $p$ be the smallest prime divisor of $n$ (such a prime exists because $n > 1$ ).

Note that $p \neq 2$ because RHS $=2^n+1 \equiv 1 \pmod 2$

So $p \mid n^2 \mid 2^n+1 \implies p \mid 2^n+1 \iff 2^n+1 \equiv 0 \pmod p \implies 2^n \equiv -1 \pmod p \implies$ $\boxed{2^{2n} \equiv 1 \pmod p}$
Also by Fermat's Little Theorem we have that : $\boxed{2^{p-1} \equiv 1 \pmod p} (\because p \neq 2 \implies gcd(2,p)=1)$
Hence $2^{\gcd(2n,p-1)} \equiv 1 \pmod p$
Note that since $p$ is the smallest prime divisor of $n$ we get that $\gcd(n,p-1)=1$ so $\gcd(2n,p-1)=1 \vee 2$

Note that since $2n \equiv 0 \pmod 2$ and $p-1 \equiv 0 \pmod 2$ we get:

$\gcd(2n,p-1)=2 \implies 2^2 \equiv 1 \pmod p \implies p \mid 3 \implies p=3 (\because p$ -prime)

Hence $3=p \mid n \implies n=3k \implies 9k^2 \mid 8^k+1$

Now suppose that $k >1$ so simmilarly as before let $q$ be the smallest prime divisor of $k$ hence we find that $8^{\gcd(2k,q-1)} \equiv 1 \pmod q$
$\gcd(2k,q-1)=2 \implies 8^2 \equiv 1 \pmod q \implies q \mid 63=7 \cdot 3^2 \implies q=3 \vee q=7$ however if $q=7$ then:
$7=q \mid 9k^2 \mid 8^k+1 \implies 7 \mid 8^k+1$ but RHS $=8^k+1 \equiv 1+1=2 \pmod 7$ so $7 \nmid RHS \rightarrow \leftarrow$

So $3=q \mid k \implies k=3 \ell \implies$ $81 \ell^2 \mid 8^{3 \ell}+1$
Now by taking $\nu_3$ we get:

$4+2 \cdot \nu_3(\ell)=\nu_3(81 \ell^2) \leq \nu_3(8^{3 \ell}+1) \stackrel{LTE}{=} \nu_3(8+1)+\nu_3(3 \cdot \ell)=3+\nu_3(\ell) \implies 4+2 \cdot \nu_3(\ell) \leq 3+\nu_3(\ell) \implies$ $\nu_3(\ell) \leq -1 \rightarrow \leftarrow$
Hence there doesn't exist a prime $q$ such that $q \mid k$ hence $k=1 \implies n=3$ which obviously works $\blacksquare$