= Solution
Write $\alpha=\sqrt[n]{m}$ and $\zeta=e^{2\pi i/n}$. The $n$ <field embeddings> of $L$ into $\mathbb C$ send $\alpha$ to $\zeta^j\alpha$, so the <field trace> is
$$
\operatorname{Tr}_{L/\mathbb Q}(\alpha^r)
=\alpha^r\sum_{j=0}^{n-1}\zeta^{jr}
=\begin{cases}
n,&r=0,\\
nm,&r=n,\\
0,&1\leq r\leq2n-2,\ r\ne n.
\end{cases}
$$
For the <trace pairing> $T(x,y)=\operatorname{Tr}_{L/\mathbb Q}(xy)$, the basis $\beta_i=\alpha^i$ therefore has <dual basis>
$$
\boxed{\beta_0^*=\frac1n,
\qquad
\beta_i^*=\frac{\alpha^{n-i}}{nm}\quad(1\leq i\leq n-1)}.
$$
Every element of $\mathbb Z[\alpha]$ is an <algebraic integer>. If $x\in\mathcal O_L$ and $z\in\mathbb Z[\alpha]$, then $xz$ is an algebraic integer and its trace is an <integer>. Thus $x$ belongs to the <trace-dual lattice> of $\mathbb Z[\alpha]$, which is generated over $\mathbb Z$ by the displayed dual basis. Since $1/n=m/(nm)$, every such linear combination lies in $(nm)^{-1}\mathbb Z[\alpha]$. Hence
$$
\boxed{\mathcal O_L\subseteq\frac1{nm}\mathbb Z[\sqrt[n]{m}]}.
$$
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