Let's determine all integers $n>1$ for which $\frac{2^{n}+1}{n^{2}}$ is also an integer.
$n=3$ is clearly a solution, since $2^{3}+1=9$ is divisible by $3^{2}=9$. We will show that no other solution exists. For this, write $n$ in the form $3^{k} \cdot d$ (where $k \geq 0$ and $d$ are coprime). Then

$$
2^{n}+1=2^{3^{k} d}+1=\left(2^{d}+1\right) \prod_{i=0}^{k-1}\left(2^{2 \cdot 3^{i} d}-2^{3^{i} d}+1\right)
$$

Since for every odd integer $t$, $2^{2 t}-2^{t}+1$ is divisible by 3 but not by 9, the $k$-factor product on the right-hand side of (1) is divisible by exactly the $k$-th power of 3.

If the number in the above form is a solution, i.e., $n^{2}=\left(3^{k} d\right)^{2}$ divides $2^{n}+1$, then the "missing" factors of 3 must be contained in the first factor of the product in (1):

$$
3^{k} \mid 2^{d}+1
$$

which is only possible for $k=0$ or 1, since $(d, 3)=1$ implies $9 \nmid 2^{d}+1$.

We will show that the value of $d$ can only be 1. Suppose $d>1$ and let the smallest prime divisor of $d$ be $p$, which is at least 5, and $(p-1, d)=1$.

From the condition,

$$
p \mid 2^{n}+1
$$

and by Fermat's little theorem,

$$
p \mid 2^{p-1}-1
$$

follows. From the congruences

$$
2^{2 n} \equiv 1(\bmod p)
$$

and

$$
2^{p-1} \equiv 1(\bmod p)
$$

it follows that

$$
2^{m} \equiv 1(\bmod p)
$$

where $m=(2 n, p-1)$.

Since $2 n \mid 2 \cdot 3 d$ and $(p-1, d)=1$, it must be that $m \mid 6$, so according to (3), the prime number $p$ must be a prime factor greater than 3 of one of the numbers $1,3,7,63$, which is only possible if $p=7$. This, however, is impossible, because according to (2), in that case $7 \mid 2^{n}+1$, which does not hold for any positive integer $n$. Therefore, $d$ indeed cannot be greater than 1.

The conditions of the problem can thus only be satisfied by numbers $n=3^{k} d$ where $k \leq 1$ and $d=1$. If $k=0$, then $n=1$, and if $k=1$, then $n=3$.