I've heard it conjectured that a finitely presentable group $G$ is hyperbolic if it satisfies the following two conditions.


1. $G$ contains no subgroup isomorphic to a Baumslag-Solitar group $BS(n,m)$ (including $BS(1,1) \cong \mathbb{Z}^2$).
2. $G$ is rationally of finite type in the sense that all the groups $H\_k(G;\mathbb{Q})$ are finite-dimensional and $H\_k(G;\mathbb{Q})=0$ for $k \gg 0$.


Question : Can someone tell me an example of a finitely presentable group that satisfies $1$ but but not $2$? All the examples of finitely presentable groups I know of that don't satisfy $2$ actually have plenty of copies of $\mathbb{Z}^2$ in them.



Nonhyperbolic groups that contain no free abelian groups or Baumslag-Solitar groups
Noel Brady's finitely presented non-hyperbolic group embeds in a hyperbolic group (and hence satisfies (1)), but has infinitely generated third integral homology. 


I'm guessing, but don't remember with certainty, that the rational third homology is infinitely generated. 


Noel Brady, Branched coverings of cubical complexes and subgroups of hyperbolic groups, J.
London Math. Soc. (2) 60 (1999), no. 2, 461–480. MR MR1724853


