Proof by induction $\gcd(2^n-1,2^m-1)=2^{\gcd(n,m)}-1$

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It is asked to perform a proof by induction over a variable $k$, which is $k=m+n$ and to use a given equation: $\gcd(a,b)=\gcd(a+b,b)=\gcd(ac+b,a)$, which might help throughout the proof-writing. Usually the induction proofs with one variable are carried out this way: assume for a $k$, that something is true. Show that something for $k+1$. Here, in this question, it is asked to assume for $k-1$ and to show for $k$ ("a shifted-index-induction"). This is one difficulty. The second difficulty is that we have to deal with the two variables $n$ and $m$ instead of just one variable as usual...

I am not clear about, how to perform the induction proof over the variable $k = m+n$...

Any help is appreciated.

Best Regards,

Ahmed Hossam

PS This question ist not a duplicate of that question, because non of the methods shown there is a proof by induction over $m+n$ and the question here comes with a helping equation $\gcd(a,b)=\gcd(a+b,b)=\gcd(ac+b,a)$, that also should be used in the proof. The question might look similar, but the method of solving, which is asked here, is not shown anywhere.