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: $(a^3-3ab^2z)^2=(a^2+b^2z)^3-(\frac{3a^2b-b^3z}{z})^2z^3$

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: $(a^3-3ab^2z)^2=(a^2+b^2z)^3-(\frac{3a^2b-b^3z}{z})^2z^3$

Edit: Of course, if you take $t=\frac{1}{x^3-y^2}, z=x^3-y^2$, you get trivial solutions, but I'm assuming you don't want those.

Really, your equation has too many variables, so writing boring solutions is easy. Are you wanting solutions that are functions of t? If so, you can rearrange my original solution above to make z a function of t, and go from there. From there, your other equations (in Addendum 2) follow by changing $t$ to $t^n$.