theory – Converse of Schreier theorem

I know that every subgroup of a free group is free (Schreier theorem).

I’m wondering that a (non-trivial) converse is true, that is, if every proper subgroup of an infinite group $G$ is free, then $G$ is free.

I think it is false but I cannot find counterexamples.

(I expect that some proper semi-direct product of the free group rank $n$ ($n geq 2$) and $mathbb{Z}/2mathbb{Z}$ is counterexample but I cannot find yet.)

Any comments would be greatly appreciated.