Suppose $G$ is a group generated by two elements $s,t$.
Suppose $H$ is a subgroup of $G$ such that $H= \langle s^k,t^k \rangle$ is a free group where $k$ is some integer not equal to $1$.
Does it imply that $G$ is also a free group?
Note: I know there are examples of non free group with a free subgroup. But in those examples, the subgroup had different generators. But I want to know about this particular case where the generators of the subgroup has some power of the generators of the mother group.
No, this is not true.
For an extreme class of counter-examples, consider hyperbolic groups, which are an extremely common* class of groups. If $G$ is a hyperbolic group then for all $g,h\in G$ there exists an integer $k\geq1$ such that $\langle g^k, h^k\rangle$ is free. So every non-free two-generated hyperbolic group gives you a counter-example. Often the promised free subgroup is free of rank two. Concrete examples include: \begin{align} \langle a, b\mid abab^2ab^3\cdots ab^n\rangle&\quad\forall\:n\gg1&(1)\\ \langle a, b\mid [a, b]^n\rangle &\quad \forall\:n>1&(2)\\ \langle a, b\mid a^n, [a, b]\rangle&\quad\forall\:n>0&(3)\\ \langle a, b\mid a^p, b^q, (ab)^r\rangle&\quad\forall\:\frac1p+\frac1q+\frac1r<1&(4) \end{align} In (1) and (2) there exists some $k$ such that $\langle a^k, b^k\rangle\cong F_2$. This is the "generic" case, as it happens if both generators have infinite order and don't commute (and a generic group is in fact "non-elementary" torsion-free hyperbolic). Note that the group in (2) is not torsion-free, although the generators have infinite order.
The result still holds for generators of finite order, it just won't be the free group of rank $2$ that you get. In (3), $\langle a^n, b^n\rangle\cong\mathbb{Z}$ which is the free group of rank $1$. In (4), $\langle a^{pq}, b^{pq}\rangle$ is trivial, which is free of rank $0$.
To save myself duplicating work: references to proofs are here. Theorem 5.3.E of Gromov's "Essay" is actually a more powerful result, covering multiple generators.
*Both in the sense that they crop up everywhere, and that they are "generic" within the class of finitely presented groups.