In my previous answer, we used a fact that an invertible ideal is projective and a fact that a finitely generated projective module over a local ring is free.
Here is a proof without using these facts.
Lemma 1
Let $A$ be a Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Let $K$ be the field of fractions of $A$.
Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} ⊂ A\}$.
Then $\mathfrak{m}^{-1} \neq A$.
Proof:
Let $a \neq 0$ be an element of $\mathfrak{m}$.
By the assumption, Supp$(A/aA) = \{\mathfrak{m}\}$.
Since Ass$(A/aA) \subset$ Supp($A/aA)$, Ass$(A/aA) = \{\mathfrak{m}\}$.
Hence there exists $b \in A$ such that $b \in A - aA$ and $\mathfrak{m}b \subset aA$.
Since $\mathfrak{m}(b/a) \subset A$, $b/a \in \mathfrak{m}^{-1}$.
Since $b \in A - aA$, $b/a \in K - A$.
QED
Lemma 1.5
Let $A$ be an integral domain.
Let $K$ be the field of fractions of $A$.
Let $M \neq 0$ be a finitely generated $A$-submodule of $K$.
Let $x \in K$ be such that $xM \subset M$.
Then $x$ is integral over $A$.
Proof:
Let $\omega_1,\dots,\omega_n$ be generators of $M$ over $A$.
Let $x\omega_i = \sum_j a_{i,j} \omega_j$.
Then $x$ is a root of the characteristic polynomial of the matrix $(a_{ij})$.
QED
Lemma 2
Let $A$ be an integrally closed Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $\mathfrak{m}$ is invertible.
Proof:
Let $K$ be the field of fractions of $A$.
Let $a \neq 0$ be an element of $\mathfrak{m}$.
Let $\mathfrak{m}^{-1} = \{x \in K; x\mathfrak{m} \subset A\}$.
Since $\mathfrak{m} \subset \mathfrak{m}\mathfrak{m}^{-1} \subset A$,
$\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$ or $\mathfrak{m}\mathfrak{m}^{-1} = A$.
Suppose $\mathfrak{m}\mathfrak{m}^{-1} = \mathfrak{m}$.
Since $\mathfrak{m}$ is finitely generated, every element of $\mathfrak{m}^{-1}$ is integral over $A$ by Lemma 1.5.
Since $A$ is integrally closed, $\mathfrak{m}^{-1} \subset A$.
This is a contradiction by Lemma 1.
Hence $\mathfrak{m}\mathfrak{m}^{-1} = A$.
QED
Lemma 3
Let $A$ be a Noetherian local domain.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $\bigcap_n \mathfrak{m}^n = 0$.
Proof:
Let $I = \bigcap_n \mathfrak{m}^n$.
Suppose $I \neq 0$.
Since dim$(A/I) = 0$, $A/I$ is an Artinian ring.
Hence there exists $n$ such that $\mathfrak{m}^n \subset I$.
Since $I \subset \mathfrak{m}^n$, $I = \mathfrak{m}^n$.
Since $I \subset \mathfrak{m}^{n+1}$, $\mathfrak{m}^n = \mathfrak{m}^{n+1}$.
By Nakayama's lemma, $\mathfrak{m}^n = 0$.
Hence $I = 0$.
This is a contradiction.
QED
Lemma 4
Let $A$ be an integrally closed Noetherian local.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Let $I$ be a non-zero ideal of $A$ such that $I \neq A$.
Then $I = \mathfrak{m}^n$ for some integer $n > 0$.
Proof:
By Lemma 3, there exists $n > 0$ such that $I \subset \mathfrak{m}^n$ and I is not contained in $\mathfrak{m}^{n+1}$.
By Lemma 2, $\mathfrak{m}$ is invertible.
Since $I \subset \mathfrak{m}^n$, $I\mathfrak{m}^{-n} \subset A$.
Suppose $I\mathfrak{m}^{-n} \neq A$.
Then $I\mathfrak{m}^{-n} \subset \mathfrak{m}$.
Hence $I \subset \mathfrak{m}^{n+1}$.
This is a contradiction.
Hence $I\mathfrak{m}^{-n} = A$.
Hence $I = \mathfrak{m}^n$.
QED
Theorem
Let $A$ be an integrally closed Noetherian local.
Suppose its maximal ideal $\mathfrak{m}$ is the unique non-zero-prime ideal.
Then $A$ is a discrete valuation ring.
Proof:
By Nakayama's lemma, $\mathfrak{m} \neq \mathfrak{m}^2$.
Let $x \in \mathfrak{m} - \mathfrak{m}^2$.
By Lemma 4, $xA = \mathfrak{m}$.
Let $I$ be a non-zero ideal of $A$ such that $I \neq A$.
By Lemma 4, $I = \mathfrak{m}^n$.
Hence $I$ is principal.
Hence $A$ is a discrete valuation ring.
QED
Best Answer
One of the things that you want in Dedeking rings is that you have nice ideal factorisation properties. In particular, multiplication by a nonzero ideal should be injective.
Suppose $R$ is a commutative ring. Let us look at an element $x/y$ of its field of fractions who happens to be a solution of a monic polynomial equation :
Let $x,y \in R$ and $a_k \in R$ for $k=0 \ldots n-1$ and suppose the equation $x^n = \sum a_k x^k y^{n-k}$ is true.
Now consider $I = (x,y)$, the ideal generated by $x$ and $y$. Then, $I^n = (x^n,x^{n-1}y, \ldots, xy^{n-1},y^n)$. But, because of that equation, the first generator is superfluous, and so $I^n = (x^{n-1}y,\ldots, xy^{n-1},y^n)$. If you look at this ideal carefully, you get $I^n = (y)I^{n-1}$.
Now, if you want your ideals to have nice factorization properties, and if $I$ is nonzero you would want $(y) = I$. Since $I = (x,y)$, this is equivalent to $x \in (y)$, so that there exists a $z \in R$ such that $x=yz$, or also that $x/y \in R$.
Therefore if you want nice ideal factorisations, you need $R$ to be integrally closed.
(checking that a few ideals factor as they should is also what I do when I want to check if a ring is integrally closed)
Also, wikipedia lists various alternative definitions, some of which place the focus on ideal factorization :
"every proper ideal factors into primes"
"every nonzero fractional ideal is invertible"
Those may be more to your liking.