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?
 A: 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,
$$(4 x^3 + 15 x^2y - 6 x y^2 - 8y^3)^2 + 
    11(x^3 - 3 x^2 y - 6 x y^2)^2 = (3x^2 + 2x y + 4y^2)^3$$
A: For $k=5$ and $a=(4mn-t^2)$
${n}^{5}\,{\left( p\,s\,t+m\,{s}^{2}+n\,{p}^{2}\right) }^{5}=\left( 4\,m\,n-{t}^{2}\right)\cdot\,{\left( -{s}^{5}\,{t}^{4}-5\,n\,p\,{s}^{4}\,{t}^{3}+\left( 3\,m\,n\,{s}^{5}-10\,{n}^{2}\,{p}^{2}\,{s}^{3}\right) \,{t}^{2}+\left( 10\,m\,{n}^{2}\,p\,{s}^{4}-10\,{n}^{3}\,{p}^{3}\,{s}^{2}\right) \,t-{m}^{2}\,{n}^{2}\,{s}^{5}+10\,m\,{n}^{3}\,{p}^{2}\,{s}^{3}-5\,{n}^{4}\,{p}^{4}\,s\right) }^{2}/4+{\left( s\,t+2\,n\,p\right)}^{2}\cdot \,{\left( {s}^{4}\,{t}^{4}+3\,n\,p\,{s}^{3}\,{t}^{3}-5\,m\,n\,{s}^{4}\,{t}^{2}+4\,{n}^{2}\,{p}^{2}\,{s}^{2}\,{t}^{2}-10\,m\,{n}^{2}\,p\,{s}^{3}\,t+2\,{n}^{3}\,{p}^{3}\,s\,t+5\,{m}^{2}\,{n}^{2}\,{s}^{4}\\-10\,m\,{n}^{3}\,{p}^{2}\,{s}^{2}+{n}^{4}\,{p}^{4}\right) }^{2}/4$ 
A: One approach is to write the General formula.  For the equation.
$$x^2+ay^2=z^{k}$$
The formula looks for simplicity without coprime solutions.  
$$x=(p^2-as^2)(p^2+as^2)^{k-1}$$
$$y=2ps(p^2+as^2)^{k-1}$$
$$z=(p^2+as^2)^2$$
After substituting numbers  $p,s - $  to reduce common divisor.
$x,y - $ in such a number. $d^k$
$z - $ in such a number. $d^2$
