<s>[INST] Positive integer solutions of $x^2+21y^2=z^4 $ Can one find all positive integer solutions of
$$x^2+21y^2=z^4 ?$$
I am not sure if this is possible. I just saw this problem and this problem came to my mind.
 [/INST] This is an old post, but I believe a clarification to the other answer might be useful. The equation,
$$x^2+dy^2 = z^k\tag{1}$$
for $k = 2$ has a complete solution in terms of the single formula,
$$((p^2-dq^2)u)^2+d(2pqu)^2=((p^2+dq^2)u)^2\tag{2}$$
where $u$ is a scaling factor. However, the situation when $k>2$ is different. One can factor $(1)$ over $\sqrt{-d}$ and find formulas for all positive integer $k$. For example,
$$\big((p^3-3dpq^2)u^3\big)^2 + d((3p^2q-dq^3)u^3\big)^2 = ((p^2+dq^2)u^2)^3\tag{3}$$
But Pepin found that additional solutions can be given, like,
$$(13u^3+60u^2v-168uv^2-144v^3)^2 + 47(u^3-12u^2v-24uv^2+16v^3)^2 = 2^3(3u^2+2uv+16v^2)^3\tag{4}$$
not covered by $(3)$. For $k=4$, the impulse is to apply the complete formula $(2)$ onto itself,
$$p = (r^2-ds^2)v$$
$$q = (2rs)v$$
to get,
$$\big((r^4 - 6 d r^2 s^2 + d^2 s^4) u v^2\big)^2 + d\big(4 r s (r^2 - d s^2) u v^2\big)^2 = (r^2 + d s^2)^4 u^2 v^4\tag{5}$$
For the RHS to be a 4th power, one has to make the assumption that $u$ in the original triple $(2)$ is a square, hence it loses its generality. For example, for $d=21$, there is no rational ${r,s,u,v}$ that yields,
$$170^2 + 21\cdot51^2 = 17^4\\
17^2\big(10^2 +21\cdot3^2 = 17^2\big)
\tag{6}$$
This instead has the form,
$$\big((p^2 - d q^2) u^2 z\big)^2 + d\big(2 p q u^2 z\big)^2 = (u z)^4\tag{7}$$
where $z = p^2 + d q^2$.  Notice that the terms of $(5)$ can be co-prime if $u=v=1$, but $(7)$ is never co-prime. 
In summary, I am not sure if $(5)$ and $(7)$ cover all cases of $k=4$. (Compare it to $k = 2$, which has only one formula whether terms are co-prime or not.)
</s>