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 $g\in G-K$ 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.

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 $g\in G-K$ 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.

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 $K< \langle K, g\rangle <G$ 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.

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 $g\in G-K$ 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.