This is a minor tweak of my answer to an earlier post by user952949. That one asked for a proof that if $a$ and $b$ are relatively prime and odd, then $\gcd(2^a+1,2^b+1)=3$.
A very useful fact: If $a$ and $b$ are relatively prime, there exist integers $x$ and $y$ such that $ax+by=1$.
We can arrange for $x$ to be $\ge 0$ and $y$ to be $\le 0$. So there are non-negative integers $u$ and $v$ such that $au=bv+1$. Thus $2^{au}=2^{bv+1}=2\cdot 2^{bv}. \qquad\text{(Equation 1)}$
Let $m$ be any common divisor of $2^a+1$ and $2^b+1$. Then $2^a \equiv -1\pmod{m}$ and $2^b\equiv -1 \pmod m$. It follows that $2^{au} =(2^a)^u \equiv (-1)^u \pmod{m} \qquad\text{and}\qquad 2^{bv} =(2^b)^v \equiv (-1)^v \pmod{m}.$ From Equation $1$, we conclude that $(-1)^u \equiv 2\cdot (-1)^v\pmod{m}.$ If $(-1)^u=(-1)^v$, we find that $1\equiv 0\pmod{m}$, so the only positive common divisor of $2^a+1$ and $2^b+1$ is $1$. If $(-1)^u=-(-1)^v$, we find that $3\equiv 0\pmod{m}$.
If we only know that $3\equiv 0\pmod{m}$, then all we can say is that any common divisor of $2^a+1$ and $2^b+1$ divides $3$.
We need to rule out the possibility that $3$ divides both of $2^a+1$ and $2^b+1$. This is easy. One of $a$ or $b$ is even, say $a$. Then $2^a\equiv 1 \pmod{3}$, so $2^a+1$ is not divisible by $3$.