The solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.
If $p \mid 7^p - 2^p$, then $p \mid 7 - 2 = 5$ by the Fermat little theorem. This implies $p = 5$. If we further have $q \mid 7^q - 2^q$, then $q = 5$ for the same reason. So $p = q = 5$ is a solution. If $q \ne 5$, then $q \mid 7^p - 2^p = 16775 = 5^2 \times 11 \times 61$. Therefore, $q = 11, 61$. By symmetry $(p, q) = (11, 5), (61, 5)$ are also solutions.

Now we may assume $p, q \ne 5$. Then we must have $p \mid 7^q - 2^q$ and $q \mid 7^p - 2^p$. WLOG assume $q \ge p$. Consider $p \mid 7^q - 2^q$. Note that $(q, p - 1) = 1$ since $q$ is a prime and $q > p - 1$. Thus, there exists $a \in \mathbb{Z}^+$ such that $aq \equiv 1 \pmod{p-1}$. By the Fermat little theorem, we have
$$7 \equiv 7^{aq} = (7^q)^a \equiv (2^q)^a = 2^{aq} \equiv 2 \pmod{p}.$$
This forces $p = 5$, which is a contradiction.

Therefore, the only solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.