Note that both $2^a+1$ and $2^b+1$ are odd, so any common divisor is odd.
Now, if $d$ divides $2^a+1$ and $2^b+1$, then it also divides $(2^a+1)-(2^b+1) = 2^a-2^b$, hence it divides $2^{\min(a,b)}(2^{|a-b|}- 1)$. But since $d$ is odd, then $d$ divides $2^{|a-b|}-1$. That is, if, say, $a\geq b$, then $d|2^{a}-1,2^b-1 \Longleftrightarrow d|2^a-1, 2^b-1, 2^{a-b}-1.$ But if $d$ divides both $2^{a-b}-1$ and $2^b-1$, then it divides $2^b(2^{a-b}-1) + (2^b-1) = 2^a-1.$ So: $d|2^a-1,2^b-1 \Longleftrightarrow d|2^b-1, 2^{a-b}-1.$ We are assuming $a\geq b$; be careful with that assumption; more generally, we have: $\text{For }a,b\text{ odd, }\gcd(2^a-1,2^b-1) = \gcd(2^{\min(a,b)}-1, 2^{|a-b|}-1).$
This looks very much like the gcd identity $\gcd(x,y) = \gcd(y,x-y),$ which is very useful. Maybe you can make it work for you?