 Solution

The assumed group is a Tarski monster group. For any $g\ne 1$, its cyclic subgroup $⟨g⟩$ must be proper. Otherwise $G$ would be infinite cyclic, and $⟨g^2⟩$ would be a proper nontrivial infinite subgroup, contrary to the hypothesis. Hence $⟨g⟩$ has order $p$.

Choose $h\notin ⟨g⟩$, possible because $G$ is infinite. The subgroup $⟨g,h⟩$ contains $⟨g⟩$ strictly. If it were proper, it would have order $p$ and could not strictly contain the order-$p$ subgroup $⟨g⟩$. ThereforeIn particular, $G$ is a finitely generated group, generated by two elements.