When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?

If $k=z$, there are no solution by Fermat's Last Theorem.
When $k=1$ a solution can be $x=1$, $y=2$, $z = 3$, because $1^{3} + 2^{3} = 3^{2}$
Another $k=38$ have a solution can be $x =3$, $y=5$, $z=2$ and there are lots of $k$ can be 
solution of that equation, but now I don't know any technique for count how many or what $k$  give a solution.
EDIT:
I found the following solution, I think it isn't the best but:
\begin{equation}
x^{3}  + y^{3} = (x+y)[(x-y)^{2}+xy]
\end{equation}
we can think that $(x+y) = z^{2}$ and $k =[(x-y)^{2}+xy]$.
now using $xy = \frac{1}{4}[(x+y)^2 - (x-y)^{2}]$
we can found $3$ solutions, 
$1) k =[(x-y)^{2}+xy] \iff x+y$ is a square;
$2) k =[\frac{3}{2}(x-y)^{2}+\frac{(x+y)^2}{2}] \iff \frac{x+y}{2}$ is a square;
$3) k =[3(x-y)^{2}+(x+y)^2] \iff \frac{x+y}{4}$ is a square;
