Example 7 (20th Irish Mathematical Olympiad) Find all prime numbers $p, q$ such that $p \mid (q+6)$ and $q \mid (p+7)$.
If $p=2$, from $p \mid(q+6)$, we get $q=2$. At this time, $q \chi(p+7)$. Therefore, $p \neq 2$.
If $q=2$, from $p \mid(q+6)$, we get $p=2$.
But $q \chi(p+7)$, which is a contradiction. Therefore, $q \neq 2$.
Thus, $p$ and $q$ are both odd primes.
Furthermore, $p+7$ is an even number.
Since $q \mid(p+7)$, we have,
$q \leqslant \frac{p+7}{2} \leqslant \frac{q+6+7}{2}$.
Hence, $q \leqslant 13$.
We use enumeration to discuss the cases $q=3,5,7,11,13$.
It is easy to see that the only solution is $q=13, p=19$.