Let $α$ be the generator of $m$, i.e. $m = (α)$.
Hint: First try to prove for any nonzero nonunit $x ∈ R$ that $(x) = (α^n)$ for some $n ∈ ℕ$, then use the fact that any ideal $I ≠ 0$ is finitely generated to conclude what you want to show.
I’ve already done that, but now realized you maybe wanted to do this yourself. But I will leave below what I already did, in case you want to take a peek.
Let $x ∈ R$ be nonzero nonunit, i.e. $(x) ≠ R$ and $(x) ≠ 0$. Then there’s a maximal $n ∈ ℕ$ such that $x ∈ m^n$ (because $x$ has to lie in the only maximal ideal $m$ at least and $\bigcap_{n ∈ ℕ} m^n = 0$). Write $x = rα^n$. Now $r$ cannot be in $m$, or else $n$ wouldn’t be maximal. Therefore $r ∈ R\setminus m = R^×$ and so $(x) = (α^n)$.
Since $R$ is noetherian, you can write any nontrivial ideal $I ≠ 0$ using nonzero generators $x_1, …, x_s ∈ R$ as $I = (x_1,…,x_s)$. Do the above argument for the generators: $I = (α^{n_1}, …, α^{n_s})$. Take $n = \min \{n_1, …, n_s\}$, then $I = (α^n) = m^n$.
$\mathbb C[X,Y]_{(X,Y)}$ is a local ring which is not a valuation ring. (In a valuation ring the ideals are linearly ordered; in our example the ideals $(X)$ and $(Y)$ are not contained one in another.)
Btw, fields are valuation rings.
Best Answer
$$R=\Bbb{Z}[x,x/p,x/p^2,\ldots,y,y/p,y/p^2,\ldots]$$ Then $(p)=(p,x,x/p,x/p^2,\ldots,y,y/p,y/p^2,\ldots)$ is a principal maximal ideal and the unique one of $R_{(p)}$.
In $R_{(p)}\subset \Bbb{Q}[x,y]_{(x,y)}$,
$(x)\not\subset (y)$ and $(y)\not\subset (x)$.