A number theory problem I saw, related to prime factors

Question

Prove that there are infinitely many prime factors of numbers of the form $2^{3^k}+1$.

Answer 1

HINT

For $N=3^k$, consider $2^{3N}+1=(2^N+1)(2^{2N}-2^N+1)$.

Let $C$ be a common factor of $2^N+1$ and $2^{2N}-2^N+1$.

Then, modulo $C$, $2^N\equiv -1$ and $2^{2N}-2^N+1\equiv 3$.

Therefore $2^N+1$ and $2^{2N}-2^N+1$ have gcd $3$.

Can you now see how to complete the proof?