Here is a proof that uses only the dynamics of the action of a hyperbolic group on its boundary.
Every hyperbolic group $G$ acts as a uniform convergence group on its Gromov boundary $\partial G$. This means that the action of $G$ on the space of ordered triples of distinct points of $\partial G$ is properly discontinuous and cocompact. For an elegant, simple proof of this fact, refer to Proposition 1.13 of Bowditch's article Convergence groups and configuration spaces.
Any subgroup $H$ of a convergence group whose limit set has at least three points must always contain a nonabelian free subgroup, by the Tits Alternative, see Theorem 2U of Tukia's paper Convergence groups and Gromov's metric hyperbolic spaces.
It follows from the Tits Alternative that any infinite torsion subgroup $H$ of a hyperbolic group must have a limit set with at most two points. Since subgroups with empty limit set are finite, and subgroups with two-point limit set are virtually cyclic, the only possibility is that there exists a point $p$ in $\partial G$ fixed by every element of the infinite torsion group $H$.
However, in a uniform convergence group action, no point can be fixed by infinitely many non-loxodromic elements. This is a key step in the proof, and a short argument can be found in Theorem 3A of Tukia's article Conical limit points and uniform convergence groups.