Write $\displaystyle A_m = 2^{2^m} + 1$ and $\displaystyle A_n = 2^{2^n}+1$ and assume WLOG that m>n.

Let $\displaystyle k = m - n$ so that $\displaystyle m = n + k$. Now we have:

$\displaystyle A_m = \left( 2^{2^m}-1 \right) +2 = \left( 2^{2^n \cdot 2^k} - 1 \right) + 2 = \left( 2^{2 \cdot 2^n} \right)^{2^{k-1}} - 1 + 2$

Let $\displaystyle b = 2^{2^n}, \ a = 2^{2 \cdot 2^n}$ so that $\displaystyle a = b^2$. Rewrite $\displaystyle A_m$:

$\displaystyle A_m = \left( a^{2^{k-1}} -1 \right) + 2 = (a-1) \left( 1 + a + a^2 + \ldots + a^{2^{k-1} -1} \right) + 2$

So if we write for simplicity's sake $\displaystyle d = \left( 1 + a + a^2 + \ldots + a^{2^{k-1} -1} \right)$ we get

$\displaystyle A_m = (a-1)d + 2 = (b^2 - 1)d + 2 = (b+1)(b-1)d + 2 = \left( 2^{2^n} + 1 \right) \left( b-1 \right) d + 2$

Therefore $\displaystyle A_m = A_n(b-1)d + 2$

And so $\displaystyle gcd(A_m, A_n) \ | \ 2$, but since they are both odd, so is their greatest common divisor, and thus $\displaystyle gcd(A_m , A_n) =1$