= Solution
For a finite extension $E/K$ of <number fields>, the <inverse different> is the <fractional ideal>
$$
\mathfrak D_{E/K}^{-1}=\{x\in E:\operatorname{Tr}_{E/K}(x\mathcal O_E)\subseteq\mathcal O_K\}.
$$
The <different ideal> is its inverse. The <trace pairing> is nondegenerate because extensions of <number fields> are separable, so this definition gives a full <fractional ideal>. The inclusion $\mathcal O_E\subseteq\mathfrak D_{E/K}^{-1}$ shows that $\mathfrak D_{E/K}$ is integral.
Write its <prime ideal factorization> as $\mathfrak D_{E/K}=\prod_{\mathfrak P}\mathfrak P^{d_{\mathfrak P}}$. For $\mathfrak P$ above $\mathfrak p$, with <ramification index> $e_{\mathfrak P}$ and residue characteristic $\ell$, the <different exponent and tame ramification> theorem says
$$
\boxed{d_{\mathfrak P}\geq e_{\mathfrak P}-1,\qquad d_{\mathfrak P}=e_{\mathfrak P}-1\ \Longleftrightarrow\ \ell\nmid e_{\mathfrak P}.}
$$
The equivalence uses the separability of finite residue-field extensions. In particular, $d_{\mathfrak P}=0$ precisely at unramified <prime ideals>; in the wild case $d_{\mathfrak P}\geq e_{\mathfrak P}$. This theorem does not require a <Galois extension>. The <relative discriminant> is the <norm of the different ideal>:
$$
\mathfrak d_{E/K}=N_{E/K}(\mathfrak D_{E/K}).
$$
For $K=\mathbb Q$, this gives $|D_E|=N(\mathfrak D_{E/\mathbb Q})$.
Now take $\alpha^3=m$ and $K=\mathbb Q(\alpha)$. For any <prime number> $q\mid m$, the <square-free integer> hypothesis makes $X^3-m$ an <Eisenstein polynomial> at $q$. It is therefore irreducible, and $K$ is a <pure cubic number field> of degree $3$. Since $\alpha$ is an <algebraic integer>, $A=\mathbb Z[\alpha]$ is an <order in a number field>. The <discriminant of elements of a number field> for its basis $1,\alpha,\alpha^2$ is
$$
\operatorname{disc}(1,\alpha,\alpha^2)=\operatorname{disc}(X^3-m)=-27m^2.
$$
For example, the resultant of $X^3-m$ and $3X^2$ is $27m^2$, and the degree-three sign in the <polynomial discriminant> is negative. If $I=[\mathcal O_K:A]$, the <discriminant-index formula for an integral lattice> gives
$$
-27m^2=I^2D_K.
$$
Only <prime numbers> dividing $3m$ can therefore divide $I$.
For $q\mid m$, the <Eisenstein polynomial> gives a <totally ramified extension> of $\mathbb Q_q$ of degree $3$. Thus $K$ has a unique <prime ideal> $\mathfrak P$ above $q$, with <ramification index> $3$ and <residue-field degree> $1$. Since $q\ne3$, this is <tame ramification>, and the <different exponent and tame ramification> theorem gives $d_{\mathfrak P}=2$. Hence $v_q(D_K)=2$. Comparing with $v_q(-27m^2)=2$ in the <discriminant-index formula for an integral lattice> yields $v_q(I)=0$.
At $3$, use the <shifted Eisenstein polynomial> of $\beta=\alpha-m$:
$$
(T+m)^3-m=T^3+3mT^2+3m^2T+(m^3-m).
$$
Because $3\nmid m$, the constant term is divisible by $3$. Moreover,
$$
9\mid m^3-m\ \Longleftrightarrow\ m\equiv\pm1\pmod9
$$
when $3\nmid m$: the factors $m-1$ and $m+1$ cannot both be divisible by $3$. The hypotheses therefore give $v_3(m^3-m)=1$, so the translated polynomial is an <Eisenstein polynomial> at $3$. The resulting completion is a <totally ramified extension> of degree $3$, with <residue-field degree> $1$, and its <ramification index> is divisible by the residue characteristic. Thus its <different exponent> is at least $3$.
It follows that $v_3(D_K)\geq3$. But $v_3(-27m^2)=3$, so
$$
3=2v_3(I)+v_3(D_K)
$$
forces $v_3(I)=0$ and $v_3(D_K)=3$. No <prime number> divides $I$. We have proved the <integral basis of a nonexceptional pure cubic field>:
$$
\boxed{\mathcal O_K=\mathbb Z[\alpha],\qquad \{1,\alpha,\alpha^2\}\text{ is an integral basis},\qquad D_K=-27m^2.}
$$
Using local <Eisenstein polynomials> here only establishes the local <ramification indices>; it does not assume that $\mathbb Z[\alpha]$ is already the full <ring of integers of a number field>.
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