It is a theorem of Selberg that a lattice $\Gamma$ in a linear group has a torsion-free subgroup of finite index. Page 64 in 'Introduction to Arithmetic Groups' by Dave Morris asserts these can be realized as follows. Since $\Gamma$ is finitely generated, $\Gamma \subseteq \mathrm{SL}_k(R)$ where $R = \mathbb{Z}[a_1, \ldots,a_n]$ for some $a_1, \ldots, a_n \in \mathbb{C}$. If $\mathfrak{m}$ is a maximal ideal in $R$ then the kernel of reduction modulo $\mathfrak{m}$ is torsion-free and has finite index in $\Gamma$ since $\mathrm{SL}_k(R/\mathfrak{m})$ is finite. For the purposes of this question let $k=2$.

In the case $\Gamma = \mathrm{SL}_2(\mathbb{Z})$ doing this gives the principal congruence subgroups, which are actually free for levels $\geq 3$. So my question is, when do lattices in $\mathrm{SL}_2(\mathbb{C})$ have a free subgroup of finite index? Relatedly, for which commutative rings $R$ does $\mathrm{SL}_2(R)$ have a free subgroup of finite index?