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?

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 $(x,y,z)=(\pm 3k^3,-2k^2,\pm k)$.
