Is there a bound $B$ such that every 2-generator subgroup $G = \langle a, b \rangle \le {\rm GL}(2,\mathbb{Z})$ whose generators do not satisfy a relation of length $\leq B$ is free?

If it exists, such bound must be at least 18, as the example $$ G = \left< \left( \begin{array}{rr} 5 & 4 \\\ -1 & -1 \end{array} \right), \left( \begin{array}{rr} 6 & 1 \\\ -1 & 0 \end{array} \right) \right> $$ shows: the shortest relation satisfied by the generators $a$ and $b$ is $a^{-2}b(ab^{-1})^3a^2b^{-1}(a^{-1}b)^3 = 1$.

Remarks:

Obviously the question can be generalized to $m$-generator subgroups of ${\rm GL}(n,\mathbb{Z})$.

The crystallographic restriction gives a positive answer to case $m = 1$ of the above generalization, thus our case $m = n = 2$ is the minimal case which is not covered.

By the Tits alternative, a subgroup of ${\rm GL}(n,\mathbb{Z})$ has either a free subgroup or a solvable subgroup of finite index.