<s>[INST] What are the integer solutions to $z^2-y^2z+x^3=0$? The question is to describe ALL integer solutions to the equation in the title. Of course, polynomial parametrization of all solutions would be ideal, but answers in many other formats are possible. For example, answer to famous Markoff equation $x^2+y^2+z^2=3xyz$ is given by Markoff tree. See also this previous question Solve in integers: $y(x^2+1)=z^2+1$ for some other examples of formats of acceptable answers. In general, just give as nice description of the integer solution set as you can.
If we consider the equation as quadratic in $z$ and its solutions are $z_1,z_2$, then $z_1+z_2=y^2$ while $z_1z_2=x^3$, so the question is equivalent to describing all pairs of integers such that their sum is a perfect square while their product is a perfect cube.
An additional motivation is that, together with a similar equation $xz^2-y^2z+x^2=0$, this equation is the smallest $3$-monomial equation for which I do not know how to describe all integer solutions. Here, the "smallest" is in the sense of question What is the smallest unsolved Diophantine equation?, see also Can you solve the listed smallest open Diophantine equations? .
 [/INST] Since the equation is NOT homogeneous, it is trivial to find
infinite families of solutions with $g=\gcd(x,y)>1$. For instance,
choose any integers $a$ and $b$, set $c=ab^2-a^2$, so multiplying
by $c^8$ gives $(ac^4)^2-ac^4(bc^2)^2+c^9=0$, giving the trivial
solutions $(x,y,z)=(c^3,bc^2,ac^4)$, but there are many other ways to
find ``trivial'' solutions.
If you assume $g=\gcd(x,y)=1$, the solution is pretty standard:
since we have a monic second degree equation in $z$, there exist
integer solutions if and only if the discriminant is a square, giving
the auxiliary equation $y^4-4x^3=d^2$, rewritten as $((y^2-d)/2)((y^2+d)/2)=x^3$,
and since the factors are coprime, they are both cubes, say $a^3$ and $b^3$,
so $x=ab$, $y^2=a^3+b^3$, hence $z=b^3$. The equation in $y$ is a standard
superFermat equation of elliptic type, which is entirely parametrized by the
following three parametrizations (see for instance my GTM 240 chapter 14 for
a proof), where $s$ and $t$ are coprime integers, and any solution belongs
to one and only one parametrization:
$$(a,b,y)=(s(s+2t)(s^2-2ts+4t^2),-4t(s-t)(s^2+ts+t^2),\pm(s^2-2ts-2t^2))$$
with $s$ odd and $s\not\equiv t\pmod3$,
$$(a,b,y)=(s^4-4ts^3-6t^2s^2-4t^3s+t^4,2(s^4+2ts^3+2t^3s+t^4),3(s^2-t^2)(s^4+2s^3t+\
6s^2t^2+2st^3+t^4))$$
with $s\not\equiv t\pmod{2}$ and $s\not\equiv t\pmod 3$,
$$(a,b,y)=(-3s^4+6t^2s^2+t^4,3s^4+6t^2s^2-t^4,6st(3s^4+t^4))$$
with $s\not\equiv t\pmod{2}$ and $3\nmid t$.
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