TITLE: Divisibility of $2^x-1 $
QUESTION [6 upvotes]: This is one of the first questions on round 1 of the Korean Math Olympiad
Find all natural numbers x, y such that $xy=2^x-1$
I "know" that the only solution is  $(x,y)=(1,1)$, but I cannot find a proof of this

REPLY [2 votes]: Lemma. If $a\not\equiv1,a^n\equiv1$ mod $m$ then $(n,\varphi(m))>1$.
Let $p\mid x$ be the smallest prime divisor. Now $p>2$, and $2^x\equiv1$ mod $p$ implies $(x,p-1)>1$ ...