The finiteness criterium $F$ under quasi-isometry A group $G$ is defined to have $F$ if there exists a finite $K(G,1)$.
This property is clearly not invariant under quasi-isometry as one can see from the trivial group and $\mathbb{Z}_2$.
My question: Is there also a torsion-free counterexample? Meaning are there two torsion-free, quasi-isometric groups of which one has $F$ and theother one not?
 A: This is not an answer but a summary of what I know about this issue:


*

*The property $F_n$ is QI invariant (the groups here are not assumed to be torsion free). You can find a proof either in section 18.2 of 


R.Geoghan, "Topological methods in group theory" 
or in chapter 6 of my book with Cornelia Drutu. 
It is quite possible that some version of the arguments in these proofs will handle your question, it is worth checking. 


*If a group $G$ has type $FP_n$ and is $n$-dimensional (here and below everything is over ${\mathbb Z}$) then $G$ has type $FP$. In particular, if a group has type $F_\infty$ and has finite cohomological dimension then it has type $FP$. 

*If a finitely presented group $G$ has type $FL$ then there is a finite $K(G,1)$. This is in Brown's book "Cohomology of groups", Ch. 8.8.

*There are no known examples of finitely presented groups which are of type FP but not of type FL.  

*Putting it all together and assuming that FP=FL, the positive answer to your question would follow from:
Suppose that $G$ admits a finite $K(G,1)$ and $H$ is torsion free quasiisometric to $G$. Then $H$ has finite cohomological dimension.  
This is quite likely to be true, since cohomological dimension over ${\mathbb Q}$ is QI invariant (R.Sauer, "Homological invariants and quasi-isometry", GAFA, 2006). 
