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

Question

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?

Answer 1

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.

Answer 2

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.