It is easy to see $n \le 2$ works. Consider $n \ge 3$. By zsigmondy's theorem, there is some prime $p$ such that $p| x^k+y^k$ but $p \nmid x+y$ with special cases( checking them still holds $n \le 2$) . Therefore any $k$ with odd prime factor , we can factor $x^k+y^k=(x+y)(stuff)$, which has at least $2$ prime divisors, contradiction. Hence $k$ is power of $2$. But it is impossible since $ x^2 \equiv 1 \mod 3$ , $ y^2 \equiv 1 \mod 3$.

When $n \leq 2$ we see that $\exists x,y,k$ satisfied. Consider $n \geq 3$. Easy to see $k$ is odd, because when $k$ is even we infer $3 \mid x,y$. We have $x+y \mid x^k+y^k=3^n$, therefore $x+y=3^q$ Using LTE theorem we have $n=v_3(x^k+y^k)=v_3(x+y)+v_3(k)=q+v_3(k)$ $\Rightarrow v_3(k)=n-q \Rightarrow k \geq 3^{n-q}$ Since $x+y$ is odd, WLOG $x>y$ We have $x^{k-1}-x^{k-2}y+x^{k-3}y^2-x^{k-4}y^3+...-xy^{k-2}+y^{k-1}>y^{k-1}$ If $y=1$ then $x^k=3^n-1$. Easy to see there is no solution for this. Therefore $y \geq 2$. Then $3^n=x^k+y^k=(x+y)(x^{k-1}-x^{k-2}y+x^{k-3}y^2-x^{k-4}y^3+...-xy^{k-2}+y^{k-1})>2^{k-1}\cdot 3^q$ $\Rightarrow 2^{k-1}>3^{n-q}$ But we have $2^{k-1} \geq k \geq 3^{n-q}$, therefore we have contradiction. So, only $n=0,n=1,n=2$ are satisfied.

uhhhhhhhh

$0, 2$ works but $1$ doesn't work. Assume $n \geq 3$; Then $3 \mid x+y$, since they are relatively prime $3 \nmid x$ and $3 \nmid y$. If $k$ is even then $3 \nmid x^k + y^k$, contradiction. If $k$ is odd; then by zsigmondy's theorem (since $3 \mid x+y$) there exist a prime $p \neq 3$ which divides $x^k + y^k = 3^n$, contradiction again. So only solutions are $n=0$ and $n=2$.