Let $m,n \in \Bbb{N}$ and $\gcd(m,n)=1$, now prove $$\gcd\left(2^m +1 ,2^n +1\right)=1$$ and show it is necessary that $\gcd(m,n)=1$.
I think that for second part of my question we can consider $m=6 ,n=2 ,\gcd(6,2)=2$ and further $2^6 +1 =65 ,2^2 +1 =5$ and $\gcd(65,5)=5$.
The second part is false. $\gcd\left(2^4+1,2^2+1\right)=1$, but $\gcd(4,2)=2\neq 1$.
The first part is true if and only if $m,n$ are not both odd.
Fact: $\gcd\left(2^m+1,2^n+1\right)\mid 2^{\gcd(m,n)}+1$ (proof below).
If $\gcd(m,n)=1$, then $\gcd\left(2^m+1,2^n+1\right)\mid 3$.
If $m,n$ are both odd, $\gcd=3$, otherwise $\gcd=1$.
Proof: We want to prove that if $p^{\alpha}\mid 2^m+1, 2^n+1$, then $p^{\alpha}\mid 2^{\gcd(m,n)}+1$.
We know $p^{\alpha}\mid 2^{2m}- 1, 2^{2n}-1$, so by this lemma:
$$p^{\alpha}\mid 2^{2\gcd(m,n)}- 1\iff p^\alpha\mid \left(2^{\gcd(m,n)}+1\right)\left(2^{\gcd(m,n)}-1\right)$$
We also know $p^{\alpha}\nmid 2^{\gcd(m,n)}-1$, because otherwise
$$p^{\alpha}\mid \left(2^{\gcd(m,n)}-1\right)\left(2^{m-\gcd(m,n)}+2^{m-\gcd(m,n)-1}+\cdots+1\right)=2^m-1,$$
contradiction. Also $\gcd\left(2^{\gcd(m,n)}+1,2^{\gcd(m,n)}-1\right)=1$. Therefore:
$$p^{\alpha}\mid 2^{\gcd(m,n)}+1$$
