Find all pairs of primes $p,q$ such that $pq \mid 2^p +2^q$ Find all pairs of primes $p,q$ such that $pq \mid 2^p +2^q$.
My attempt  : When either one of them is $2$ then easy case checking gives me set of solutions. But what happens when neither of them is $2$? 
 A: A slight tune up to the argument presented in Prometheus' answer (currently deleted, but hopefully soon fixed and undeleted) proves that there are no solutions such that $p$ and $q$ are both odd primes.
Lemma. Assume that $p$ is a prime factor of $2^{2^ts}+1$ where $t,s$ are positive integers, $s$ odd. Then
$$
p\equiv1\pmod{2^{t+1}}.
$$
Proof. Let $m$ be the order of the residue class of $2$ in $\Bbb{Z}_p^*$.
Because $p\mid 2^{2^ts}+1$, we have $2^{2^ts}\equiv-1\pmod p$, and consequently $2^{2^{t+1}s}\equiv1\pmod p$. This means that $m\mid 2^{t+1}s$ but $m\nmid 2^ts$. Therefore we can conclude that $2^{t+1}\mid m$. But by Lagrange's theorem $m\mid p-1$, so the claim follows.
The rest is easy. Without loss of generality we can assume that $2<p<q$ (the case $p=q$ is obviously impossible). Write $q-p=2^ts$, $s$ odd. The assumption is that
$$
pq\mid 2^p+2^q=2^p(2^{q-p}+1).
$$
This implies that both $p$ and $q$ should be factors of $2^{q-p}+1$. But the Lemma then implies that
$$
p\equiv q\equiv 1\pmod{2^{t+1}}.
$$
This means that $q-p$ is divisible by $2^{t+1}$ contradicting the fact that $s$ is an odd integer by design.
