Let $x,y \in \mathbb Z$ such that both $x$ and $y$ are not factor of $3$. Prove that $\forall k \in \mathbb N$ there exists $(x , y)$ that $x^2 + 2y^2 = 3^k$.
I know that $x^2 + 2y^2$ is divisible by $3$ always. But how can we prove that there exists a solution $\forall k \in \mathbb N$? Thank you!