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.



Free subgroups of $\mathrm{GL}(2,\mathbb{Z})$
In Olʹshanskiĭ, A. Yu.; Sapir, M. V. [On $F\_k$-like groups](http://mi.mathnet.ru/al/v48/i2/p245). (Russian) Algebra Logika 48 (2009), no. 2, 245--257, 284, 286--287; 
[translation](https://doi.org/10.1007/s10469-009-9044-2) in Algebra Logic 48 (2009), no. 2, 140–146, 
we proved (Theorem 2) that every non-virtually cyclic hyperbolic group is $F\_k$-like, that is for every $m$ it has a generating set consisting of $k$ elements which do not satisfy any relation of length $\le m$. The minimal $k$ that works is 1 plus the number of generators of the group. In particular, for $GL\_2(Z)$, $k=3$. 


For the group $\mathbb{Z}\*\mathbb{Z}\_2$ (as in Yves' comment), $k=2$ also works: one can adapt the proof of Theorem 1 of the paper (we prove, in particular, that every group with 2-generated presentation satisfying $C'(1/6)$ is $F\_2$-like). 


