More specifically, can you show that it is greater than $1$ unless $x$ is $1, 2$, or $5$?
I was trying to solve $3^x=2y^2+1$ in the positive integers, and since $3=(1+\sqrt{-2})(1-\sqrt{-2})$ and $2y^2+1=(1-y\sqrt{-2})(1+y\sqrt{-2})$
where $\gcd(1+\sqrt{-2},1-\sqrt{-2})=\gcd(1+y\sqrt{-2}, 1-y\sqrt{-2})=1$, I figured the real part of $(1+\sqrt{-2})^x$ would have to be $\pm 1$, but I can't seem to get further from there. I may just be having a brain fart here, but can anyone help me out?
Let $a_m$ be the real part of $1+i\sqrt{2})^m$, so $$ a_m = \frac{(1+i\sqrt{2})^m}{2} + \frac{(1-i\sqrt{2})^m}{2}. $$ Then $a_m = 2a_{m-1} - 3a_{m-2}$, with $a_0 = 1$ and $a_1 = 1$.
You ask about proving $|a_m| \to \infty$ as $m \to \infty$. This is equivalent to proving $a_m$ has any particular value a finite number of times. Proving this in either form is not easy! A solution is Theorem 1.1 here, which uses $p$-adic analysis, mainly Strassmann’s theorem, and the link to the Diophantine equation you ask about is Theorem A.1 near the end.
A question about determining all $m$ such that $a_m = \pm 1$ was posted on this site $10$ years ago here. The solution I outlined there uses the $2$-adic field $\mathbf Q_2(\sqrt{-2})$.
