No positive integer solution for $x^2 + y^3 = z^6$? I am thinking about the following problem:

Is there any solution to $x^2 + y^3 = z^6$, where $x, y, z$ are positive integers?

I searched all possible solutions for $1 \leq z \leq 1000$ and there were no solution. So I think there is no solution, but I cannot prove it.
Here, one possible approach is to transform the equation to $y^3 = (z^3 - x)(z^3 + x)$. I tried in this way but cannot reach the proof.
The main reason is because we cannot assume that $x, y, z$ are coprime without loss of generality.
Of course I think there are other approach of proving. Can you tell me how to prove it?
 A: This is the proof referenced in the comments above.
The only solutions of this equation in nonzero integers are $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$, where $k$ is nonzero.
Let $(x,y,z)$ be a solution with minimal $|z|$. from equation we derive that $(z^3-x)(z^3+x)=y^3$. Obviously $\gcd (x,y)=\gcd(y,z)=\gcd(z,x)=1$. Thus there are coprime integers $a,b$ such that $z^3-x=a^3$ and $z^3+x=b^3$ and therefore $a^3+b^3=2z^3$, or there are coprime integers $a,b$ such that $z^3-x=2a^3$ and $z^3+x=4b^3$ and therefore $z^3+(-a)^3=2b^3$.
In both cases we are left with an equation of type $a^3+b^3=2z^3$ with $\gcd(a,b)=1$. let $|z|$ be minimal (assume that $|z|\ne0$). $a$ 
and $b$ must be odd. now substitute $a=u+v$ and 
$b=u-v$, where $u$ and $v$ are coprime integers and from different parity. then $u (u^2+3v^2)= z^3$ and so there are integers $tr,s$ such that $u = r^3$ and $u^2+3v^2= s^3$.
From this, we have $(u+\sqrt{-3}v)(u-\sqrt{-3}v)=s^3$. treating this equation in $\Bbb Z[\sqrt{-3}]$, you can reach a equation of the form $a_0^3+b_0^3=2z_0^3$, where $|z_0|<|z|$, unless $a=b$ (or $|z|=0$) which yields the solutions $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$.
A: This is an example of the Generalized Fermat Equation, viz;
$$x^p+y^q=z^r$$
For $(x, y, z) \in \mathbb{Z}/\{0\}$
Here, $(p, q, r)$ in the literature is known as the signature of the equation. In particular the behaviour of primitive solutions depends fundamentally upon the size of the quantity
$$\sigma(p, q, r) = \frac{1}{p}+\frac{1}{q}+\frac{1}{r}$$
Notice that $\chi = \sigma(p, q, r)-1$ is Euler Characteristic and that then depending on whether $\sigma(p, q, r) > 1$, $\sigma(p, q, r) = 1$ or $\sigma(p, q, r)<1$ one can classify as spherical, parabolic and hyperbolic.
This case is clearly of parabolic nature and since we have $\sigma(p,q,r) = 1$, then, up to reordering,
$$(p,q,r) = (2,6,3),(2,4,4),(4,4,2),(3,3,3), (2,3,6)$$
The only primitive non-trivial solution comes from the signature $(p,q,r) = (2,3,6)$, also corresponds to the Catalan solution $3^2 −2^3 = 1$.
