Let $x$ and $y$ be two permutations of $\mathbb{Z}^2$ defined as follows. The permutation $x$ sends $(n,0)$ to $(n+1,0)$ and fixes all else while $y$ sends $(0,n)$ to $(0,n+1)$ and fixes all else. Is the group generated by $x$ and $y$ amenable?

I do know that the group does not contain a copy of the free group on two generators, so it is very likely to be amenable. I also know that if $y$ is defined, instead, by sending $(n,m)$ to $(n,m+1)$ then the group generated by $x$ and $y$ is amenable, in fact, it is a solvable extension of a locally finite group.