Parametrization of $x^2+ay^2=z^k$, where $\gcd(x,y,z)=1$ $x,y,z$ be three coprime integers, $a \in \mathbb{Z}>0$ and $k$ an odd integer. How do I find all the non-trivial solutions of the diophantine equation? $$x^2+ay^2=z^k$$
Does the method which consists in assuming that: $$x^2+ay^2=(p^2+aq^2)^k$$ (where $p,q$ are two coprime integers) provide the general parametrization?

No, there is no single method. For example, we have, 
$$(p^3-3dpq^2)^2 + d(3p^2q-dq^3)^2 = (p^2+dq^2)^3$$
but it is no longer complete. In particular, for $d=47$, Pepin found there is no rational $p,q$ that corresponds to the solution,
$$(13u^3+30u^2v-42uv^2-18v^3)^2 + 47(u^3-6u^2v-6uv^2+2v^3)^2 = 2^3(3u^2+uv+4v^2)^3$$
Edit: 
Using Jagy's $x^2+2xy+4y^2$, we have,
$$(4 x^3 + 15 x^2y - 6 x y^2 - 8y^3)^2 + 
    11(x^3 - 3 x^2 y - 6 x y^2)^2 = (3x^2 + 2x y + 4y^2)^3$$
