Prove that two distinct number of the form $a^{2^{n}} + 1$ and $a^{2^{m}} + 1$ are relatively prime if a is even and have $gcd=2$ if a is odd
My attempt:
If $a$ is even, let $a = 2^{s}k$ for some integers $k, s$
Then, $a^{2^{n}} + 1 = 2^{2^{n}s}\times k^{2^n} + 1$ and $a^{2^{m}} + 1 = 2^{2^{m}s}\times k^{2^m} + 1$ To prove that they're relatively prime, we need to show that their gcd = 1. And I was stuck here, how could I prove that gcd of two numbers is $1$?
A hint would be sufficient. Thanks.