Assume there exists infinitely many prime numbers $l$ such that $2^l-1$ is NOT a prime, prove that ther exists infinitely many pairs $(p,q)$ of DISTINCT prime numbers $p \neq q$ s.t.
$p\mid (2^{q-1}-1)$ and $q\mid (2^{p-1}-1)?$
*So I walked into my professor's office hour and he suggest me to use the following fact $\gcd(2^a-1,2^b-1) = 2^{\gcd(a,b)} -1 $, and since here $\gcd(2^{p-1}-1,2^l-1) \ge p \gt 1$, we eventually will have $l \mid p-1$; simialrly for $q$ case, and I finally figure out this problem. Thank you to everyone for helping.