Find the pairs of prime numbers $(p, q)$ such that $pq$ divides $3^{p}+3^{q}$.
Let $(p, q)$ be a potential solution. If $p \geqslant 5$, then $0 \equiv 3^{p}+3^{q} \equiv 3+3^{q} \equiv 3\left(3^{q-1}+1\right)(\bmod p)$, hence $3^{q-1} \equiv -1 \equiv 1 \equiv 3^{2(q-1)}(\bmod p)$. We deduce that $\omega_{p}(3)$ divides $2(q-1)$ but not $q-1$, so $v_{2}\left(\omega_{p}(3)\right)>v_{2}(q-1)$. Since $\omega_{p}(3)$ divides $\varphi(p)=p-1$, this means that $v_{2}(p-1)>v_{2}(q-1)$. Therefore, it is impossible to have both $p \geqslant 5$ and $q \geqslant 5$.

Thus, we distinguish several cases, assuming without loss of generality that $p \geqslant q$:

- if $p \leqslant 3$ and $q \leqslant 3$, we manually verify that $(3,2)$ and $(3,3)$ are solutions, while $(2,2)$ is not;
- if $q=2$ and $p \geqslant 5$, then $3^{p}+3^{q} \equiv 3+3^{2} \equiv 12(\bmod p)$, so $p$ does not divide $3^{p}+3^{q}$;
- if $q=3$ and $p \geqslant 5$, then $3^{p}+3^{q} \equiv 3+3^{3} \equiv 30(\bmod p)$, so $(p, q)=(5,3)$ is the only solution.

In conclusion, the solutions are the pairs $(2,3),(3,2),(3,3),(3,5)$ and $(5,3)$.