Can $3^x=2y^2-1$ be solved over the natural numbers?

Question

In the equation $3^x=2y^2-1$, $x$, $y$ are natural numbers. I found $x=1$ or $2$ (mod $4$), and $y^2=1$ or $4$ (mod $120$) but I even don't know if the number of solutions is infinite. Is there a way to find the solution of this indeterminate equation?

Answer 1

Take both sides of the equation modulo $3$. If $x\geq 1$ then $2y^2\equiv 1$ modulo $3$, but if $y\equiv 0$, then $2y^2\equiv 0$, and if $y\equiv\pm 1$, $2y^2\equiv 2(\pm1)^2\equiv2$. So there are no solutions in this case. We must have $x=0$, thus $y=1$.

Answer 2

You can even solve this question using Mathematical Induction. However, answer is also correct and easy.

And its solution is x=0 and y=1.