Example 6 (CMO-24 Test) Find all prime pairs $(p, q)$ such that $p q \mid\left(5^{p}+5^{q}\right)$.

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If $2 \mid p q$, without loss of generality, let $p=2$, then $2 q\left|\left(5^{2}+5^{q}\right) \Rightarrow q\right|\left(5^{q}+25\right)$.
By Fermat's Little Theorem, we know $q \mid\left(5^{q}-5\right)$, thus $q \mid 30$, i.e., $q=2,3,5$.
It is easy to verify that the prime pair $(2,2)$ does not meet the requirements, while $(2,3),(2,5)$ do.
If $p q$ is odd and $5 \mid p q$, without loss of generality, let $p=5$, then
$5 q\left|\left(5^{5}+5^{q}\right) \Rightarrow q\right|\left(5^{q-1}+625\right)$.
When $q=5$, the prime pair $(5,5)$ meets the requirements.
When $q \neq 5$, by Fermat's Little Theorem, we have $q \mid\left(5^{q-1}-1\right)$, hence $q \mid 626$. Since $q$ is an odd prime, and the only odd prime factor of 626 is 313, thus, $q=313$.
Upon verification, the prime pair $(5,313)$ meets the requirements.
If $p, q$ are neither 2 nor 5, then $p q \mid\left(5^{p-1}+5^{q-1}\right)$, so
$5^{p-1}+5^{q-1} \equiv 0(\bmod p)$.
By Fermat's Little Theorem,
$5^{p-1} \equiv(\bmod p)$.
From equations (1) and (2), we get
$5^{q-1} \equiv-1(\bmod p)$.
Let $p-1=2^{k}(2 r-1), q-1=2^{l}(2 s-1)(k, l, r, s$ be positive integers).
If $k \leqslant l$, then by equations (2) and (3), it is easy to see
$$
\begin{aligned}
1 & =1^{2^{l-k}(2 s-1)} \equiv\left(5^{p-1}\right)^{2^{l-k}(2 s-1)} \\
& \equiv 5^{2^{l}(2 r-1)(2 s-1)}=\left(5^{q-1}\right)^{2 r-1} \\
& \equiv(-1)^{2 r-1} \equiv-1(\bmod p),
\end{aligned}
$$

which contradicts $p \neq 2$. Therefore, $k>l$.
Similarly, $k<l$ leads to a contradiction.
Thus, there are no $(p, q)$ that meet the requirements in this case.
In summary, the prime pairs $(p, q)$ that satisfy the requirements are $(2,3),(3,2),(2,5),(5,2),(5,5),(5$,
$313),(313,5)$.