Example 6 Find all pairs of primes $(p, q)$ such that
$$
p q \mid\left(5^{p}+5^{q}\right) .{ }^{[2]}
$$
(2009, China Mathematical Olympiad)
(1) If $\min \{p, q\}=2$, without loss of generality, let $p=2$, then to make $2 q l\left(5^{2}+5^{q}\right)=25+5^{q}$, it is necessary that
$$
q \mid\left(25+5^{q}\right) \text {. }
$$

Since $5^{q}+25=5^{q}-5+30$, by Fermat's Little Theorem, we have
$$
q\left|\left(5^{q}-5\right) \Rightarrow q\right| 30 \text {. }
$$

At this point, $q=2,3,5$.
When $q=2$, $p q=2 \times 2=4 \times\left(5^{2}+5^{2}\right)$, which does not meet the requirement;

When $q=3$, $p q=2 \times 3=61\left(5^{2}+5^{3}\right)$, so $(p, q)=(2,3)$ is a solution;

When $q=5$, $p q=2 \times 5=101\left(5^{2}+5^{5}\right)$, so $(p, q)=(2,5)$ is a solution.
(2) If $\min \{p, q\}=5$, without loss of generality, let $p=5$, to make $5 q \mid\left(5^{5}+5^{q}\right)$, it is necessary that $q \mid\left(5^{q-1}+625\right)$.

When $q=5$, $p q=5 \times 5=25 \mid\left(5^{5}+5^{5}\right)$, so $(p, q)=(5,5)$ is a solution.

When $q \neq 5$, $q \perp\left(5^{q-1}+625\right)$, and by Fermat's Little Theorem, we have $q \mid\left(5^{q-1}-1\right)$, then
$$
q \mid 626 \Rightarrow q=313 \text {. }
$$

Below, we prove that $(p, q)=(5,313)$ meets the requirement.
Notice that
$$
\begin{array}{l}
p q=5 \times 313 \mid\left(5^{5}+5^{313}\right), \\
5^{5}+5^{313}=5\left(5^{4}+5^{312}\right)=5\left(5^{312}-1+626\right) .
\end{array}
$$

Since $313 \mid\left(5^{312}-1+626\right)$, so,
$$
p q=5 \times 313 \mid\left(5^{5}+5^{313}\right),
$$

so $(p, q)=(5,313)$ is a solution.
(3) If $\min \{p, q\}>5$, then to make $p q \mid \left(5^{p}+5^{q}\right)=5\left(5^{p-1}+5^{q-1}\right)$, thus,
$$
p q \mid\left(5^{p-1}+5^{q-1}\right) \text {. }
$$

By Fermat's Little Theorem, we have
$$
5^{p-1} \equiv 1(\bmod p) .
$$

From equations (1) and (2), we get
$$
5^{q-1} \equiv-1(\bmod p) \text {. }
$$

Let $p-1=2^{k}(2 r-1)$,
$$
q-1=2^{l}(2 s-1)\left(k, l, r, s \in \mathbf{N}_{+}\right) \text {. }
$$

If $k \leqslant l$, then from equations (2) and (3), we get
$$
\begin{array}{l}
1 \equiv 1^{2 l-k(2 s-1)} \equiv\left(5^{p-1}\right)^{2^{l-k}(2 r-1)} \\
\equiv 5^{2 k(2 r-1) \cdot 2^{l-k}(2 s-1)} \\
\equiv 5^{2 l(2 r-1)(2 r-1)} \equiv\left(5^{q-1}\right)^{2 r-1} \\
\equiv(-1)^{2 r-1} \equiv-1(\bmod p),
\end{array}
$$

i.e., $2 \equiv 0(\bmod p)$.
This contradicts $p \neq 2$.
Similarly, when $k \geqslant l$, it is also a contradiction.
At this point, there are no prime pairs $(p, q)$ that meet the requirement.
In summary, and considering the symmetry of $p$ and $q$, the prime pairs that satisfy the problem's requirements are
$$
\begin{aligned}
(p, q)= & (2,3),(3,2),(2,5),(5,2), \\
& (5,5),(5,313),(313,5) .
\end{aligned}
$$

There are a total of seven solutions.