Do there exist infinitely many pairs of primes $(p,q)$ such that $pq$ divides $2^{p-1}+2^{q-1}-2$? A mathematician friend gave me this question (partly as a joke) a few months ago and it has puzzled me for a long time:-

Do there exist infinitely many pairs of primes $(p,q)$ such that $pq|2^{p-1}+2^{q-1}-2$?

The equation seems so random I have been unable to make any significant progress on it (I can just derive equally horrible conditions).
I would appreciate it enormously if anyone could set me on the right track regarding this this.
 A: No one has explicitly said this yet: Robert Israel's formula always works. Let $(p,q) = (4k+1, 8k+1)$. In one direction, $2^{q-1} -1 = (2^{p-1})^2-1$ and we know $2^{p-1} \equiv 1 \mod p$ so $p | 2^{q-1}-1$. In the other direction, since $q \equiv 1 \mod 8$, we have $\left( \frac{2}{q} \right) = 1$ so $2^{(q-1)/2} \equiv 1 \mod q$ and $q | 2^{(q-1)/2}-1=2^{p-1}-1$.
A: Hint:
$p,q > 3$ 
$pq|2^{p-1}+2^{q-1}-2$
Reframe to $pq|(2^{q-1}-1)+(2^{p-1}-1)$, now you know $2^{q-1}-1 \equiv 0(\mod q)$ and similarily $2^{p-1}-1 \equiv 0(\mod p)$. 
$\dfrac{pk+qt}{pq}=s$, For some $k,t,s \in \mathbb{N}$
If we consider the case, $p$ and $q=2p-1$ (Yeah, taking the hint from Robert Israel's comment, $p \equiv 1(\mod 4)$)
$2^{p-1}-1 \equiv 0(\mod p)$ and $2^{2p-2}-1 \equiv (\mod q) \implies (2^{\frac{2p-2}{2}}-1)(2^{\frac{2p-2}{2}}+1) \equiv 0(\mod q)$
It is still not known that $(p,2p-1)$ pairs of primes are infinite.
A: See also http://www.mathsolympiad.org.nz/wp-content/uploads/2010/04/2010squad-nt-soln.pdf (last page) for an elementary solution which does not rely on Zsigmondy's theorem.
