Prove that if $n$ and $m$ are coprime then $2^n-1$ and $2^m-1$ respectively are coprime too

Question

How to prove that if $n$ and $m$ are coprime then $2^n-1$ and $2^m-1$ are coprime too? I tried with Bezout's identity without results.

Thanks

Answer 1

Show that $\gcd(2^n-1, 2^m-1) = 2^{\gcd(m, n)} - 1 = 2^1-1= 1$.