MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

I'll expand on Derek Holt's comment, which answers your question. Suppose one has a group $G$ of the type you describe, so that finitely generated subgroups are generated by $r$ elements and have exponent $n$. Consider a finitely generated subgroup $K< G$. By the restricted Burnside problem, there is a universal constant $R(r,n)$ such that $|K|\leq R(r,n)$. Now, choose the largest size subgroup $K< G$ which is finitely generated. Since $K$ is finite and $G$ is infinite, there exists $g\in G$ G-K$such that $K < \langle K, g\rangle <G$ is finitely generated, so$\langle K, g\rangle$must be finite. But since$|K|$is maximal, we have$K=\langle K,g\rangle$, so$g\in K$, a contradiction. 1 [made Community Wiki] I'll expand on Derek Holt's comment, which answers your question. Suppose one has a group$G$of the type you describe, so that finitely generated subgroups are generated by$r$elements and have exponent$n$. Consider a finitely generated subgroup$K< G$. By the restricted Burnside problem, there is a universal constant$R(r,n)$such that$|K|\leq R(r,n)$. Now, choose the largest size subgroup$K< G$which is finitely generated. Since$K$is finite, there exists$g\in G$such that$K< \langle K, g\rangle