5. Find the positive integer $n$, such that $\frac{2^{n}+1}{n^{2}}$ is an integer.
5. Solution: When $n \geqslant 4$, let $p$ be the smallest prime factor of $n$. Clearly, $p$ is odd, and $p \mid 2^{n}+1$, so $p \mid 2^{2 n}-1$.

Also, $p \mid 2^{p-1}-1$ (Fermat's Little Theorem), so $p \mid 2^{(2 n, p-1)}-1$.
Since $(n, p-1)=1$, we have $p \mid 2^{(2, p-1)}-1$, so $p \mid 3 \Rightarrow p=3$.
Let $n=3^{k} \cdot m\left(k \in \mathbf{N}_{+}\right),(3, m)=1$.
When $k \geqslant 2$, $3^{2 k}\left|2^{3^{k} \cdot m}+1 \Rightarrow 3^{k+2}\right| 2^{3^{k} \cdot m}+1$.
Also, $2^{3^{k} \cdot m}+1=(3-1)^{3^{k} \cdot m}+1$
$$
\begin{array}{l}
=\sum_{j=0}^{3^{k} \cdot m} C_{3^{k} \cdot m}^{j} \cdot 3^{j}(-1)^{3^{k} \cdot m-j}+1 \\
=\sum_{j=1}^{3^{k} \cdot m} C_{3^{k} \cdot m}^{j} \cdot 3^{j}(-1)^{3^{k} \cdot m-j} .
\end{array}
$$

For $2 \leqslant j \leqslant 3^{k} \cdot m$,
$$
C_{3^{k} \cdot m}^{j} \cdot 3^{j}(-1)^{3^{k} \cdot m-j}=(-1)^{3^{k} \cdot m-j} 3^{j} \cdot \frac{3^{k} \cdot m \cdot\left(3^{k} \cdot m-1\right) \cdots\left(3^{k} \cdot m-j+1\right)}{j!} .
$$

Thus, $L_{3}\left(C_{3^{k} \cdot m}^{j} \cdot 3^{j}(-1)^{3^{k} \cdot m-j}\right) \geqslant L_{3}\left(3^{j} \frac{3^{k} \cdot m\left(3^{k} \cdot m-1\right) \cdots\left(3^{k} \cdot m-j+1\right)}{j!}\right)$
$$
\begin{array}{l}
\geqslant j+k-\sum_{i=1}^{j}\left[\frac{j}{3^{i}}\right] \geqslant k+j-\left(\frac{j}{3}+\frac{j}{3^{2}}+\cdots+\frac{j}{3^{j}}\right) \\
=k+j-\frac{j}{3} \cdot \frac{1}{1-\frac{1}{3}}=k+\frac{j}{2} \geqslant k+1 .
\end{array}
$$

Thus, $L_{3}\left(C_{3^{k} \cdot m}^{j} \cdot 3^{j}(-1)^{3^{k} \cdot m-j}\right) \geqslant k+2$ (where $L_{p}(a)$ denotes the exponent of the prime $p$ in the standard factorization of $a$).
So, $2^{3^{k} \cdot m}+1 \equiv C_{3^{k} \cdot m}^{1} \cdot m \cdot 3\left(\bmod 3^{k+2}\right) \equiv m \cdot 3^{k+1}\left(\bmod 3^{k+2}\right) \not \equiv 0\left(\bmod 3^{k+2}\right)$.
Thus, $k=1$, so $n=3 m$ (where $m$ is odd).
Since $n \geqslant 4$, we have $m>1$. Let $q$ be a prime factor of $m$,
then $q \geqslant 5, q \mid 2^{3 m}+1 \Rightarrow q \mid 2^{6 m}-1, q \mid 2^{q-1}-1$,
so $q \mid 2^{(6 m, q-1)}-1=2^{(6, q-1)}-1$.
Thus, $q \mid 2^{6}-1 \Rightarrow q \mid 63$, and since $q>3$, we have $q=7$.
So, $7 \mid 8^{m}+1$ is a contradiction! Therefore, $n \leqslant 3$.
Upon inspection, $n=1,3$ satisfy the conditions, and $n \neq 2$.
Thus, the values of $n$ that satisfy the conditions are 1 or 3.