Infinite group with the order of abelian subgroup bounded

To complement Arturo's answer:

A famous theorem of Ph. Hall is that every infinite locally finite group has an infinite abelian subgroup.

As a consequence, an infinite group whose abelian subgroups are finite of bounded order, has an infinite finitely generated subgroup, clearly sharing the same property. It's, in particular, of finite exponent. So, there's no easy approach to your question.

Adian in 1979 proved that for odd $n\ge 665$ and all $m\ge 2$, the Burnside group $B(m,n)$ (free group of exponent $n$ on $m$ generators) is infinite and all its abelian subgroups are cyclic (hence of order $\le n$).

Note also that finitely generated groups of finite exponent can have infinite abelian subgroups: for instance, if $G$ is infinite, finitely generated of exponent $n$ then $(\mathbf{Z}/p\mathbf{Z})\wr G$ is finitely generated of exponent $pn$ and has an infinite abelian subgroup of exponent $p$.

Tarski monsters provide examples. These are (infinite) groups $G$ in which every proper subgroup $H$ is either trivial, or cyclic of order a fixed prime $p$. In particular, the only abelian subgroups are of order $1$ or $p$.

Such groups exist for every sufficiently large prime, as shown by Olshanski'i. They are 2-generated, nonabelian, simple, and a rich source of all sorts of counterexamples.