Proving that if $m>0,n>0$, and n is odd, then $\gcd(2^m+1,2^n-1)=1$.

Question

$\gcd(2^{mn}+1,2^{mn}-1)=\gcd(2,2^{mn}-1)=1$

So I think $2^{m}+1|2^{mn}+1$ when n is odd

And $2^n-1|2^{mn}-1$

So $\gcd(2^m+1,2^n-1)=1.$

But my classmate said I wrote it wrong.