If there are only finitely many primes $\equiv 1\mod 2^k$, let $P$ be their product (as always $P=1$ if there are none) and consider a prime divisor $q$ of $P^{2^{k-1}}+1$: $P^{2^{k}} \equiv 1 \mod q$, but $P^{2^{k-1}} \equiv -1 \not \equiv 1 \mod q$ gives that the order of $P \mod q$ divides $2^{k}$ but doesn't divide $2^{k-1}$, so equals $2^{k}$. Fermat's little theorem $P^{q-1} \equiv 1 \mod q$ then gives $2^{k} | q-1$, in other words $q \equiv 1 \mod 2^{k}$, contradiction ($q$ clearly doesn't divide $P$).

We can use Dirichlet's theorem on arithmetic progressions!

Drichlet theorem