Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$. 
Find all solutions in positive integers to $3^n=x^k +y^k$ where $\gcd(x,y)=1$ and $k \ge 2$.

Firstly if $k$ is even, then as $k=2t$ for $t \in \Bbb Z$ and so $x^k=(x^t)^2$ and $y^k=(y^t)^2$. These are both perfect squares and since $3^n \mid x^k +y^k$ we have that $3 \mid x^k +y^k \implies 3 \mid(x^t)^2+(y^t)^2$ but if $3 \mid a^2 +b^2$, then $3 \mid a$ and $3 \mid b$ so $3 \mid x$ and $3 \mid y$ implying that the $\gcd(x,y) \ne 1$ which is a contradiction.
Thus we can conclude that $k$ is odd. Now for odd $k$ and prime $p$ dividing $x+y$ we have that $$v_p(3^n)=v_p(x^k+y^k)=v_p(x+y)+v_p(k)$$ however as $p \mid x+y$ we have that $1=v_p(p) \le v_p(x+y)$ thus $$v_p(3^n) \ge1>0$$ so $p \mid 3 \implies p=3$.
We then have that $$v_3(3^n)=n = v_3(x+y)+v_3(k)$$ which implies that $$3^n=3^{v_3(x+y)}\cdot 3^{v_3(k)} = x^k+y^k.$$
I couldn't proceed further from here, but the solution I read stated that $$x^k+y^k = 3^{v_3(x+y)}\cdot 3^{v_3(k)} = \color{red}{(x+y)k}$$ and I have no idea where the rhs of the equality comes from?
Edit I think one has that from $n = v_3(3^n)=v_3(x+y)+v_3(k)$ we get $$3^n=3^{v_3(x+y)}\cdot 3^{v_3(k)} = x^k+y^k=(x+y)(x^{k-1}-x^{k-2}y+ \dots + y^{k-1})$$ so either $$3^{v_3(x+y)}=x+y, 3^{v_3(k)}=x^{k-1}-x^{k-2}y+ \dots + y^{k-1}$$ or vice versa.
 A: EDIT: It appears I read too fast and OP wasn't asking for a solution, but rather an explanation of an already given solution. Nevertheless, I think my solution is pretty neat and clean, so I'll leave it up.

As you've shown, $k$ must be odd. Let $p\mid k$ be prime, then
$$\frac{x^k+y^k}{x^p+y^p}=\frac{x^k-(-y)^k}{x^p-(-y)^p}\in\mathbb{Z}.$$
Hence, $x^p+y^p$ must be a power of three as well. Write
$$3^m=x^p+y^p$$
Now, note that $y$ is invertible modulo $3^m$, so
$$(x/y)^p\equiv -1\pmod {3^m}.$$
Now, $(\mathbb{Z}/3^m\mathbb{Z})^\times$ is cyclic of order $2\cdot 3^{m-1}$ and $p$ is odd. So if $p\neq 3$, we must have $x/y\equiv -1\pmod {3^m}$ and
$$x^p+y^p=3^m\mid x+y,$$
whence $x+y\ge x^p+y^p$. It follows that $x=y=1$, but this doesn't give a solution. We conclude that $p=3$ and
$$3^m=x^3+y^3.$$
Note that $x+y\mid x^3+y^3$, so there exists some positive integer $\ell$ satisfying
$$3^\ell=x+y\quad\text{and}\quad 3^{m-\ell}=\frac{x^3+y^3}{x+y}=x^2-xy+y^2.$$
This implies that
$$3^{m-\ell}=(x+y)^2-3xy=3^{2\ell}-3xy\equiv 3\pmod 9,$$
so $m-\ell=1$ and $3=x^2-xy+y^2\ge 2xy-xy=xy$. Since $3\nmid xy$, this gives only the options
$$(x,y)=(1,1),\quad (x,y)=(2,1),\quad (x,y)=(1,2).$$
Indeed we find that those last two are solutions.
