How to show in a clean way that $z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$ is a torus? How to show in a clean way that the zero-locus of $$z^4 + (x^2 + y^2 - 1)(2x^2 + 3y^2-1) = 0$$ is a torus? 

If one happens to know some Morse theory the following is another possible solution:
Consider the function $\Phi:\mathbb{R^3}\rightarrow\mathbb{R}$ defined by $$\Phi(x,y,z)=z^4+(x^2+y^2-1)(2x^2+3y^2-1).$$ You want to prove that $\Phi^{-1}(0)=:S$ is a torus. Computing $d_{(x,y,z)}\Phi$ you can check that $0$ is a regular value and hence $S$ is indeed a smooth surface embedded in $\mathbb{R}^3$. In particular $S$ is orientable.
Define now the function $$f:S\rightarrow\mathbb{R},\ (x,y,z)\mapsto y.$$ Using that $T_{(x,y,z)}S=\ker d_{(x,y,z)}\Phi$ it is not difficult to check that $f$ is a Morse function with critical points $(0,-1,0),(0,-\frac{1}{\sqrt{3}},0),(0,\frac{1}{\sqrt{3}},0),(0,1,0)$. Now use the Hessian of $f$ at these points to show that their indices are $0,1,1$ and $2$, respectively. Using the Morse inequalities you get that $\chi(S)=0$.
Since $S$ is an orientable surface with Euler characteristic equal to $0$, it must be a torus.
