When does $p^s = 2^t + 1$ have a solution, for $t$ odd and $p$ prime?

Question

If $p, s, t \in \mathbb{N}$, when does $$p^s = 2^t + 1$$ have a solution, for $t$ odd and $p$ prime?

Note that, when $s = 1$, we have $p \equiv 1 \pmod 4$, and therefore $t = 2^s$ for some $s \in \mathbb{N}$ (i.e., $p$ is a Fermat prime).

I do not know what happens when $s > 1$, or if it is even possible.

Answer 1

Catalan's conjecture is very far from being elementary, so I have sketched an alternative, elementary proof.

If $t>1$ then $$2^t+1=3(2^{t-1}-2^{t-2}+2^{t-3}-\cdots+1)$$ so $p=3$. Now, write the equation this way: $$3^s-1=2^t$$ or $$3^{s-1}+3^{s-2}+\cdots+1=2^{t-1}$$ Since the LHS has $s$ odd terms and the RHS is even, then $s$ is even. Let $s=2m$ and now $$(3^m+1)(3^m-1)=2^t$$ Therefore both $3^m+1$ and $3^m-1$ are powers of two, and they must be $2$ and $4$ (because no other pair of powers of two differ in two units). Thus, $m=1$, $s=2$ and $t=3$.

Answer 2

$3^2$ and $2^3$ are the only pairs of consecutive perfect power (Catalan's conjecture / Mihăilescu's theorem).

Therefore, the only solution is $(p,s,t) = (3,2,3)$.