First, neither $x$ or $y$ is divisible by $3$ because they are co-prime.

Next, $k$ cannot be even because otherwise $x^k$ and $y^k$ will be both $1\pmod{3}$, leading to $x^k+y^k$ not a power of $3$. 

If $n=0$, there will be no positive solution to $x^k + y^k=1$ given $k\ge 2$. So we exclude this situation.

When $n > 1$, $3\not\mid x$, $3\not\mid y$, and $x\mid x+y$, applying LTE (# 4802): $$n=v_3(x^k+y^k)=v_3(x+y)+v_3(k)\implies x^k+y^k=3^n=3^{v_3(x+y)}3^{v_3(k)}=(x+y)k$$

The last equation holds because neither $x+y$ nor $k$ can have any divisor other $3$. Furthermore, the left side is usually much larger than the right side. So it is possible to determine the upper boundary of these numbers.

Suppose $x>y$ without loss of generality. Dividing both side by $x+y$ leads to $$x^{k-1}-x^{k-2}y+\cdots-xy^{k-2}+y^{k-1}=k$$

or $$(x-y)(x^{k-2} + x^{k-4}y^2+\cdots+xy^{k-3})+y^{k-1}=k$$

The leftside is greater or equal to $x^{k-2}$. Hence, $x^{k-2}\le k \implies \ln{x}\le\frac{\ln{k}}{k-2}$. Recall that $k\ge 3$ and $x\ge 2$. By calculus, the right side decreasing for $k\ge 3$, so $\ln{x}\le\ln{3}$. So $x\le 3$. We already ruled out $x$ is a multiple of $3$. Hence $x=2$ and $y=1$.

In that case, $2^{k-2}>k$ already unless $k=3,4$. However, we know that $k$ is odd, leading to $k=3$. It is easier to check $x=2$, $y=1$, $k=3$ is a solution, as is $x=1$, $y=2$, $k=3$ if we relax the assumption of $x>y$.