number theory - Show that if $p$ is a prime such that $p|(2^{64}+1)$ then $p equiv 1$ (mod 128)

I'm not sure if I'm on the right track with this problem. So far I've said: $2^{64} = (2^{32})^2 \equiv -1$ (mod p). Then by Fermat's two square theorem $p = 2$ or $p \equiv 1$ (mod 4). We know $p \not = 2$ because $p|(2^{64}+1)$. Then $p \equiv 1$ (mod 4). From here I'm unsure on how to proceed.