There is a beautiful characterization of prime, radical, irreducible and primary ideals among monomial ones in $k[x_1, \dots, x_n]$:
Theorem. Let $I$ be a monomial ideal of $k[x_1, \dots, x_n]$ and let $\mathcal{B}$ be its minimal basis. Then:
- $I$ is maximal iff $\mathcal{B}=\{x_1, \dots, x_n \}$;
- $I$ is prime iff $\mathcal{B} = \{ x_{i_1}, \dots, x_{i_r} \}$;
- $I$ is radical iff $\mathcal{B}$ is made up of square-free monomials;
- $I$ is irreducible iff $\mathcal{B} = \{ x_{i_1}^{a_1}, \dots, x_{i_r}^{a_r} \}$;
- $I$ is primary iff $\mathcal{B} = \{ x_{i_1}^{a_1}, \dots, x_{i_r}^{a_r}, m_1, \dots, m_s \}$, where $m_1,\dots, m_s$ are monomials in the variables $x_{i_1}, \dots, x_{i_r}$.
So in this case it is very easy to produce a counter-example: $(x^2, y^2, xy)$. Its radical is maximal, so it is primary, but is reducible because $(x,y^2) \cap (x^2, y) = (x^2, y^2, xy)$.
As user26857 answered the question in a comment, and prefers not to post it as an answer, I'll try to write the answer myself. I think I've understood user26857's argument, but I may be wrong. So, in the lines below, everything that's true is due to user26857, and everything that's false is due to me.
The answer is Yes.
More precisely:
If $A$ is a noetherian integral domain, if $\mathfrak p_1,\dots,\mathfrak p_k$ are distinct nonzero prime ideals of $A$, and if $m$ and $n$ are distinct elements of $\mathbb N^k$, then we have $$\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}\ne\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}.$$
Proof. In the setting of the question, suppose by contradiction that we have
$$
\mathfrak p_1^{m_1}\cdots\mathfrak p_k^{m_k}=\mathfrak p_1^{n_1}\cdots\mathfrak p_k^{n_k}
$$
with $m\ne n$.
Enumerate the $\mathfrak p_i$ in such a way that each $\mathfrak p_i$ is a minimal element of the set $\{\mathfrak p_i,\dots,\mathfrak p_k\}$, and write $\mathfrak p_{ij}$ for the localization of $\mathfrak p_i$ at $\mathfrak p_j$.
For all $i$ we get
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{n_1}\cdots(\mathfrak p_{ii})^{n_i}.\quad(1)
$$
Note the following consequence of the determinant trick, or Nakayama's Lemma:
$(2)$ If $\mathfrak a$ and $\mathfrak b$ are ideals of $A$, then the equality $\mathfrak a\mathfrak b=\mathfrak b$ holds only if $\mathfrak a=(1)$ or $\mathfrak b=(0)$.
Let's prove $m_i=n_i$ by induction on $i$:
Case $i=1$: We have $(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{n_1}$ by $(1)$. If we had $m_1\ne n_1$ we could assume $m_1<n_1$, and would get
$$
(\mathfrak p_{11})^{n_1-m_1}(\mathfrak p_{11})^{m_1}=(\mathfrak p_{11})^{m_1},
$$
contradicting $(2)$.
From $i-1$ to $i$: We have
$$
(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{m_i}=(\mathfrak p_{1i})^{m_1}\cdots(\mathfrak p_{i-1,i})^{m_{i-1}}(\mathfrak p_{ii})^{n_i}.\quad(3)
$$
If we had $m_i\ne n_i$ we could assume $m_i<n_i$ and we could write $(3)$ as
$$
(\mathfrak p_{ii})^{n_i-m_i}\mathfrak b=\mathfrak b
$$
with $(\mathfrak p_{1i})^{n_i-m_i}\ne(1)$ and $\mathfrak b\ne(0)$, contradicting $(2)$. (Here $\mathfrak b$ is the left side of $(3)$, and we assume $2\le i\le k$.) $\square$
Note that the argument still works if $A$ is not noetherian, but the $\mathfrak p_i$ are finitely generated.
Best Answer
You ask about three different classes of nonzero ideals in a Dedekind domain $R$:
(a) Prime ideals.
(b) Unfactorable ideals: if $\mathfrak{p} = IJ$ then $I = R$ or $J = R$.
(c) Irreducible ideals: If $\mathfrak{p} = I \cap J$ then $I = \mathfrak{p}$ or $J = \mathfrak{p}$.
Classes (a) and (b) are the same: their equivalence follows easily from the fact that every nonzero ideal in a Dedekind domain factors uniquely as a product of primes.
Class(c) consists precisely of the prime powers $\mathfrak{p}^a$ for $a \in \mathbb{N}$. This follows from the following more general result about intersections of ideals in a Dedekind domain:
Let me know if you need help in proving this. If you know about localizations, this allows for a very easy proof: you can reduce to the case in which $R$ has exactly one nonzero prime ideal, and in this case the result is trivial.
Finally, one has the following simple result.
Proof: $\neg$ (i) $\implies$ $\neg$ (ii): The hypotheses give the existence of $x \in I \setminus \mathfrak{p}$ and $y \in J \setminus \mathfrak{p}$ such that $xy \in \mathfrak{p}$. Thus $\mathfrak{p}$ is not prime.
$\neg$ (ii) $\implies$ $\neg$ (i): If $\mathfrak{p}$ is not prime, then there are $x,y \in R \setminus \mathfrak{p}$ with $xy \in \mathfrak{p}$. Take $I = (x)$, $J = (y)$.
I believe this answers all of your questions, but let me know if I missed something.