This question already has an answer here:
If $p$ is a (obviously odd) prime divisor of $2^{2^n}+1$, wee see that $2^{2^n} = -1 \mod p$ and thus $2^{2^{n+1}} = 1 \mod p$. This shows $2^{n+1}=ord_p(2)$, in particular $n$ is uniquely determined by $p$. Thus if $p$ divides $2^{2^n}+1$ and $2^{2^m}+1$ we get $n=m$.
WLOG let $m>n$ and $m=n+c, c>0$ and if $a^{2^n}+1=r$
Now $$a^{2^m}+1=(a^{2^n})^{2^c}+1=(r-1)^{2^c}+1\equiv2\pmod r$$
$$\implies\left(a^{2^m}+1,a^{2^n}+1\right)=\left(2,a^{2^n}+1\right)$$