Complementing Misha's answer, the following is true for any non-(virtually cyclic) hyperbolic group $G$:

**Claim:** for any sufficiently large $k \in \mathbb{N}$ there is a free quasiconvex subgroup $H_k \le G$ of rank $k$ and an infinite order element $g \in G$ such that $G=\langle H_k,g \rangle$.

By sufficently large, I mean that $k \ge K-1$, where $K$ is the smallest integer such that $G$ can be generated by $K$ elements of infinite order. Thus $K=rank(G)$ if $G$ is torsion-free. It is not difficult to show that $K \le rank(G)+1$ in general.

The above claim is especially easy to prove if $G$ is torsion-free and is generated by two elements $h,g \in G$.

Indeed, since the subgroup $H_1:=\langle h \rangle$ is quasiconvex and has infinite index in $G$, by a theorem of Goulnara Arzhantseva ["On quasiconvex subgroups of word hyperbolic groups", Geometriae Dedicata, 87 (2001), 191-208.], there exists $h_2\in G\setminus\{1\}$ such that the subgroup $H_2:=\langle H,h_2\rangle\le G$ is quasiconvex and is isomorphic to the free product $H*\langle h_2\rangle$, which is evidently a free group of rank $2$. Clearly, $G=\langle H_2,g \rangle$. Now, without loss of generality, we can assume that $|G:H_2|=\infty$ (one can replace $h_2$ with its square to ensure this). Applying Arzhantseva's result again one finds $h_3 \in G\setminus\{1\}$ such that $H_3:=\langle H_2,h_3 \rangle\le G$ is free of rank $3$, quasiconvex in $G$ and has infinite index. And so on.

Of course, if $r:=rank(G) >2$ then the proof becomes more technical, as the first $r-1$ generators may not generate a free quasiconvex subgroup of infinite index. However, one can construct an argument by induction, using generalizations of Arzhantseva's theorem.