Find the integers $n$ such that there exist $x, y, k$ with $x$ and $y$ coprime and

$$
3^{n}=x^{k}+y^{k}
$$
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 mod 3 for $a$ and $b$). Suppose there exists an odd prime $p$ that divides $x+y$, then by the LTE lemma, we would have

$$
v_{p}\left(3^{n}\right)=v_{p}\left(x^{k}+y^{k}\right)=v_{p}(x+y)+v_{p}(k) \geqslant 1
$$

We deduce that $p \mid 3^{n}$, and thus $p=3$. This means that $x+y=3^{m}$, for a positive integer $m$. We notice that $n=v_{3}(k)+m$. Let's distinguish two cases:

- if $m>1$: I leave it to you to prove that $3^{a} \geqslant a+2$ and thus $v_{3}(k) \leqslant k-2$. We can assume $x>y$ and since $x+y=3^{m} \geqslant 9$, we will have $x \geqslant 5$, but also $x>\frac{3^{m}}{2}$. We now use some inequalities:

$$
x^{k}+y^{k} \geqslant x^{k} \geqslant \frac{3^{m}}{2} 5^{k-1}>3^{m} 5^{k-2} \geqslant 3^{m+v_{3}(k)}=3^{n}
$$

which is a contradiction.

- if $m=1$, then we have $x=2$ and $y=1$, and $3^{1+v_{3}(k)}=1+2^{k}$. We notice that $3^{1+v_{3}(k)} \mid k$, so $3^{1+v_{3}(k)} \leqslant k$ and $2^{k}+1 \leqslant 3 k$, and thus $k \leqslant 3$. There are not many cases left to check, and the only solution is:

$$
3^{2}=1^{3}+2^{3}
$$