Integer solutions of $x^3+y^3=z^2$ Is there any integer solution other than $(x,y,z)=(1,2,3)$ for $x^3+y^3=z^2$?

You can work in $\mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{3}}.$ 
There are two essential cases to deal with. In both cases, you can easily reduce to considering the case that $x,y,z$ are pairwise coprime ( you can afterwards multiply $x$ and $y$ by $d^{2}$ and $z$ by $d^{3}$ for some integer $d$). The case when $3$ does not divide $z$ is easier.
In that case, you can assume that $x + y = a^{2}$ for some integer $a$ and that $x + \omega y = (b + \omega c)^{2}$ for integers $b$ and $c.$ Then you can see that 
$x^{3} + y^{3} = (a(b^{2}-bc +c^{2}))^{2}.$ Note that the second equation gives
$x = b^{2}-c^{2}$ and $y = 2bc - c^{2}$ since $1 + \omega + \omega^{2} = 0.$ 
Hence we additionally need the integers $a,b,c$ to satisfy $a^{2} + 3c^{2} = (b+c)^{2},$ but this is easy to arrange.
In the (still pairwise coprime case that $3$ divides $z,$ you must have $x + y = 3^{2m-1}a^{2}$ for some integer $a$ coprime to $3$, and $x + \omega y = (1- \omega)(b + \omega c)^{2}$ for some integers $b$ and $c$ with $b+c$ not divisible by $3$.
