I think that what I ask must be true. I would need such lemma to finish the proof that for every ideal $I$ in $\mathcal{O}_K$ there exists a field extension $L$ of $K$ such that $I$ is a principal ideal in $\mathcal{O}_L$.
My proof goes like the first answer to this question: Every ideal of an algebraic number field can be principal in a suitable finite extension field
However, as I have not yet been introduced to fractional ideals, I do not see why, perhaps trivially, $I^m=J^m$ implies $I=J$, where $I$ and $J$ are both fractional ideals. But I have seen a proof of unique factorization into prime ideals for the ring of integers of a number field (I know that unique factorization into prime ideals holds for any Dedekind domain, despite not having yet learned the proof of this).
Can one prove what I ask in the question without needing fractional ideals? Or, if they are handy for this purpose, could you explain which properties of fractional ideals are used to deduce that $I^m=(a)^m$ implies $I=(a)$, for ideals in $\mathcal{O}_L$ ?