= Solution
Choose $0\ne a\in\mathfrak m$. Since the one-dimensional local domain has no nonzero prime ideal other than $\mathfrak m$, one has $\sqrt{(a)}=\mathfrak m$. Finite generation of $\mathfrak m$ therefore gives some $n\geq1$ with $\mathfrak m^n\subseteq(a)$. Choose $n$ minimal and
$$
x\in\mathfrak m^{n-1}\setminus(a).
$$
In the fraction field of $A$, put $y=x/a$. Then $y\notin A$ but
$$
y\mathfrak m\subseteq A.
$$
If $y\mathfrak m\subseteq\mathfrak m$, multiplication by $y$ would preserve the nonzero finitely generated faithful $A$-module $\mathfrak m$. The <determinant trick> would make $y$ integral over $A$, contradicting that $A$ is integrally closed and $y\notin A$. Hence some $t\in\mathfrak m$ satisfies $yt\notin\mathfrak m$. Since $yt\in A$ and $A$ is local, $yt$ is a unit.
For any $z\in\mathfrak m$, one has $yz\in A$, and therefore
$$
\frac zt=\frac{yz}{yt}\in A.
$$
Thus $\mathfrak m\subseteq(t)$, while the reverse inclusion follows from $t\in\mathfrak m$. Consequently
$$
\boxed{\mathfrak m=(t)}.
$$
This proves the <principal maximal ideal in a one-dimensional normal local domain> result.
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