Diffeomorphism from parametrized torus to set of zeros

Question

I am asked to prove that the differentiable map $$ h \colon S^1 \times S^1 \to M,~~(a_1,a_2,b_1,b_2) \mapsto (\sqrt{2}+b_1) \cdot \begin{pmatrix} a_1 \\ a_2 \\ 0 \end{pmatrix} + b_2 \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$$ is a diffeomorphism where $M$ is the set of points in $\mathbb{R}^3$ that satisfy $$ (x^2 +y^2+z^2+1)^2 = 8 (x^2 +y^2).$$ I already understood that $S^1 \times S^1$ is a torus and that $M$ also forms a torus so I intuitively understand that it has to be true. To prove the claim I tried to find an inverse map of $h$, $h^{-1}: M \to S^1 \times S^1,~~(x,y,z) \mapsto (a_1,a_2,b_1,b_2)$ but I can not find a way to "get out" $b_1$ since I only seem to know $b_2$ and the fact that $b_1^2+b_2^2 = 1$ which gives me only $b_1 = \pm \sqrt{1-b_2^2}$. I also proved that $M$ is a submanifold of $\mathbb{R}^3$ using the fact that $M = f^{-1}(0)$ for $0$ a regular value of the following map $$ f \colon \mathbb{R}^3 \to \mathbb{R},$$ $$ (x,y,z) \mapsto (x^2+y^2+z^2+1)^2 - 8(x^2+y^2).$$ Am I just not seeing the right way to find $h^{-1}$ or is there some other way of proving that $h$ is a diffeomorphism?