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Thomas
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If we rewrite the equation as $y^2+t^{2}z^3 = x^3$, and let $v=tz$, then $y^2+zv^2=x^3$.

If you factorize over $Q[\sqrt{-z}]$, then $(y+v\sqrt{-z})(y-v\sqrt{-z})=x^3 $.

Let $x=(a+b\sqrt{-z})(a-b\sqrt{-z})$, then $y+v\sqrt{-z}=(a+b\sqrt{-z})^3=(a^3-3ab^2z)+(3a^2b-b^3z)\sqrt{-z}$

Thus $y=a^3-3ab^2z, v=3a^2b-b^3z, x=a^2+b^2z$, and $t=\frac{v}{z}=\frac{3a^2b-b^3z}{z}$

We then have the general parametric solution: $(a^3-3ab^2z)^2=(a^2+b^2z)^3-(\frac{3a^2b-b^3z}{z})^2z^3$

Thomas
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