Find all positive integers $(x, y, n, k)$ such that $x$ and $y$ are coprime and such that

$$
3^{n}=x^{k}+y^{k}
$$
First, $k$ must be odd. Indeed, if $k$ were even, $x^{k}$ and $y^{k}$ would be squares, and it is easy to verify that $3\left|a^{2}+b^{2} \Longrightarrow 3\right| a$ and $3 \mid b$ (it suffices to check all possible congruences $\bmod 3$ for $a$ and $b$). If $(x, y, k) \neq(2,1,3)$, we can apply Zsigmondy's theorem, which provides a prime $p$ dividing $x^{k}+y^{k}$ but not $x+y$. Since $x+y$ divides $x^{k}+y^{k}$, this implies that $x^{k}+y^{k}$ has at least two prime divisors. The only solution is therefore $(x, y, k)=(2,1,3)$.