The modular group group $\text{PSL}_2(\mathbb{Z})$ can be written as something that is nearly a free group on two elements: $$ SL_2(\mathbb{Z}) \simeq \mathbb{Z}/2\mathbb{Z} \ast \mathbb{Z}/3\mathbb{Z} = \left\{ S,T: S^2 =I; (ST)^3 = I\right\} $$ where $S: z \mapsto - \frac{1}{z}$ and $T : z \mapsto z + 1$ are the reciprocal and shift maps. These can turn into many many things.
What happens if we adjoin $\sqrt{2}$ and look at the special linear group there? Do we still get that it is the free group on two elements? $$ SL_2\big(\mathbb{Z}[\sqrt{2}] \big) = \left\{ \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)\in M_{2 \times 2}(\,\mathbb{Z}[\sqrt{2}] \,): ad -bc=1 \right\} $$
These are not Bianchi groups because I am asking about $\sqrt{2}$ instead of $\sqrt{-2}$.
On the bright side, $\mathbb{Z}[\sqrt{2}]$ is a one of a handful of quadratic fields that are Euclidean domains, (so they are "norm-Euclidean"). Therefore, solutions to $ad-bc = 1$ are not too hard to generate with $(a,b,c,d) \in \mathbb{Z}[\sqrt{2}]^4$ but not in $\mathbb{Z}^4$.
Yes, as every arithmetic group (as mentioned by Moishe Cohen). See Soulé's paper "An introduction to arithmetic groups", Theorem 6 p13, https://hal.archives-ouvertes.fr/hal-00001348/document
He refers to a 1963 paper by Armand Borel and Raghunathan's book "Discrete subgroups of Lie groups".
$\mathrm{SL}_2(\mathbf{Z}[\sqrt{2}])$ is an irreducible, non-cocompact lattice, in $\mathrm{SL}_2(\mathbf{R})^2$. Its behavior, in many respects, is closer to that of $\mathrm{SL}_3(\mathbf{Z})$ than to that of $\mathrm{SL}_2(\mathbf{Z})$. In particular, finite presentation, using a retraction, is proved in a similar way. I guess there are also approaches in the spirit of algebraic K-theory.