Example 2 Find all integers $n>1$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer.
It is easy to conjecture that $n=3$ is the only solution that meets the requirements. Below, we prove that this is indeed the case.

The proof requires several steps. The first point is to consider the smallest prime divisor $p$ of $n$, and from $n \mid (2^n + 1)$, derive that $p=3$, as seen in Example 1 of Unit 8. Therefore, we can assume
$$n=3^m c, \quad m \geqslant 1, \quad 3 \nmid c.$$

The second step is to prove that $m=1$. From $n^2 \mid (2^n + 1)$, we get $2^{3^m c} \equiv -1 \pmod{3^{2m}}$, hence
$$2^{2 \times 3^m c} \equiv 1 \pmod{3^{2m}}.$$

If $m \geqslant 2$, then by Example 5 of Unit 8, the order of 2 modulo $3^{2m}$ is $2 \times 3^{2m-1}$, so from (2) we deduce that $2 \times 3^{2m-1} \mid 2 \times 3^m c$, i.e., $3^{m-1} \mid c$, thus $3 \mid c$ (since $m \geqslant 2$), which contradicts (1) where $3 \nmid c$. Therefore, we must have $m=1$.

The third step is to prove that $c=1$ in (1). This can be done similarly to the first step, i.e., Example 1 of Unit 8: $\square$

If $c>1$, let $q$ be the smallest prime divisor of $c$, then
$$2^{3c} \equiv -1 \pmod{q}.$$

Let $r$ be the order of 2 modulo $q$. From (3), we get $2^{6c} \equiv 1 \pmod{q}$, and also $2^{q-1} \equiv 1 \pmod{q}$, so $r \mid 6c$ and $r \mid (q-1)$, hence $r \mid (6c, q-1)$. By the selection of $q$, we have $(q-1, c)=1$, so $r \mid 6$. From $2^r \equiv 1 \pmod{q}$, we deduce that $q=3$ or 7. It is easy to see that $q=3$ is impossible; and from (3), $q=7$ is also impossible. Therefore, we must have $c=1$. Thus, $n=3$.

Note that if we first prove that $c=1$ in (1), it will not be effective. The order of the arguments here is quite important. Additionally, in the second step, $m=1$ can also be proven by comparing powers of primes:

By the binomial theorem,
$$2^n + 1 = (3-1)^n + 1 = 3n + \sum_{k=2}^n (-1)^k \binom{n}{k} 3^k.$$

Let $3^a \parallel k!$, then
$$\alpha = \sum_{l=1}^{\infty} \left[\frac{k}{3^l}\right] k + m - \frac{k}{2} \geqslant m + 1.$$

Thus, $\beta \geqslant m + 2$, so the sum on the right-hand side of (4) is divisible by $3^{m+2}$. If $m>1$, then $2m \geqslant m+2$, so from $3^{2m} \mid (2^n + 1)$ and (4), we get $3^{m+2} \mid 3n$, i.e., $3^{m+1} \mid n$, which contradicts (1). Therefore, we must have $m=1$.