Let $H$ be an infinite hyperbolic group. Can $H \rtimes \mathbb Z$ ever be hyperbolic? It is well known that hyperbolic groups cannot contain a copy of $\mathbb Z^2$. It follows that if $H$ is infinite and hyperbolic, then $G = H \times \mathbb Z$ cannot be hyperbolic, as $H$ will contain some non-torsion element which will then commute with the generator of $\mathbb Z$.
I'm curious as to what happens if we restrict this condition slightly, in particular, we have the following:

Let $H$ be an infinite hyperbolic group. Can $H \rtimes_\phi \mathbb Z$ ever be hyperbolic for any choice of $\phi$?

My instinct is that the geometry of the situation prohibits this, but I cannot quite see how to show it. Any help or examples would be appreciated.
 A: There are early examples arising in Thurston's work during the age of hyperbolic manifolds, a little before the era of hyperbolic groups: if $H = \pi_1(S)$ where $S$ is a closed, oriented surface of genus $g \ge 2$ (and so $H$ is hyperbolic), and if $\phi \in \text{Aut}(H)$ is induced by a pseudo-Anosov homeomorphism of $S$, then $H \rtimes_\phi \mathbb Z$ is the fundamental group of closed hyperbolic 3-manifold (and so $H \rtimes_\phi \mathbb Z$ is a hyperbolic group).
The actual statement of the examples above expressed in the language of hyperbolic groups was given first, as an application of Thurston's earlier work, in a paper of Bestvina and Feighn. In that same paper they produced further examples: for any free group $H$ of finite rank $\ge 2$ and for any $\phi \in \text{Aut}(H)$, if there does not exist a proper, nontrivial free factor $F < H$ and $n \ge 1$ such that $\phi^n(F)$ is conjugate to $F$, and if there does not exist an infinite cyclic subgroup $C < H$ and $n \ge 1$ such that $\phi^n(C)$ is conjugate to $C$, then $H \rtimes_\phi \mathbb Z$ is a hyperbolic group.
