Solve equation $ 1+2^x=3^y$

Find integers $x$ and $y$ such that$$ 1+2^x=3^y.$$ It is obvious that $x = y = 1$ and $x = 3, y = 2$ are solutions. I think others are not. How to show that?

Case 1: $y$ is odd. Then $3^y \equiv 3 \mod 8$. When is $2^x \equiv 2 \mod 8$?

Case 2: $y = 2 z$ is even. Then $2^x = 3^{2z} - 1 = (3^z - 1)(3^z + 1)$, so $3^z - 1$ and $3^z + 1$ are powers of $2$. What powers of $2$ differ by $2$?

By the way, this proof of a special case of Catalan's conjecture dates back to Gersonides (Levi ben Gershon) in 1343.

Clearly in this case Robert Israel's very elegant solution is perfect, but it is worth noting that what we have here is a special case of Catalan's conjecture (now a theorem, proved by Preda Mihăilescu) which states that the only solution of

$$x^a - y^b = 1$$ in integers (greater than 1) is given by $3^2-2^3 = 1$, so you are correct, there are no other solutions.