I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.
It's clear that $\mathbb{Z}[\sqrt{-3}]$ is not a Dedekind domain, as it is not integrally closed. I'm having difficulty demonstrating directly that $\mathbb{Z}[\sqrt{-3}]$ does not admit unique factorization of ideals.
Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.
Best Answer
In the ring ${\mathbf Z}[\sqrt{-3}]$, the ideal $P = (2,1+\sqrt{-3})$ is prime since it has index 2 in the ring. Note $P^2 = (4,2+2\sqrt{-3},-2 + 2\sqrt{-3}) = (4,2+2\sqrt{-3}) = (2)(2,1+\sqrt{-3}) = (2)P$, where $(2)$ is the principal ideal generated by 2 in ${\mathbf Z}[\sqrt{-3}]$. If there were unique factorization of ideals into products of prime ideals in ${\mathbf Z}[\sqrt{-3}]$ then the equation $P^2 = (2)P$ would imply $P = (2)$, which is false since $1+\sqrt{-3}$ is in $P$ but it is not in $(2)$.
In fact we have $P^2 \subset (2) \subset P$ and this can be used to prove the ideal (2) does not even admit a factorization into prime ideals, as follows. If $Q$ is a prime ideal factor of (2) then $(2) \subset Q$, so $PP = P^2 \subset Q$, which implies $P \subset Q$ (from $Q$ being prime), so $Q = P$ since $P$ is a maximal ideal in ${\mathbf Z}[\sqrt{-3}]$. If (2) is a product of prime ideals then it must be a power of $P$, and $P^n \subset P^2$ for $n \geq 2$, so the strict inclusions $P^2 \subset (2) \subset P$ imply (2) is not $P^n$ for any $n \geq 0$.
The "intuition" that the ideal $P = (2,1+\sqrt{-3})$ is the key ideal to look at here is that $P$ is the conductor ideal of the order ${\mathbf Z}[\sqrt{-3}]$. The problems with unique factorization of ideals in an order are in some sense encoded in the conductor ideal of the order. So you want to learn what a conductor ideal is and look at it in several examples. For example, the ideals $I$ in ${\mathbf Z}[\sqrt{-3}]$ which are relatively prime to $P$ (meaning $I + P$ is the unit ideal (1)) do admit unique factorization into products of prime ideals relatively prime to $P$. That illustrates why problems with unique factorization of ideals in ${\mathbf Z}[\sqrt{-3}]$ are closely tied to the ideal $P$.
If you look at the ideal notation $P' = (2,1+\sqrt{-3})$ in the larger ring ${\mathbf Z}[(1+\sqrt{-3})/2]$, which we know has unique factorization of ideals, then we don't run into any problem like above because $P' = 2(1,(1+\sqrt{-3})/2) = (2)$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$, so $P'$ is actually a principal ideal and the "paradoxical" equation $PP = (2)P$ in ${\mathbf Z}[\sqrt{-3}]$ corresponds in ${\mathbf Z}[(1+\sqrt{-3})/2]$ to the dumb equation $P'P' = P'P'$. (The ideal $P'$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$ is prime since the quotient ring mod $P'$ is a field of size 4: ${\mathbf Z}[(1+\sqrt{-3})/2]$ is isom. as a ring to ${\mathbf Z}[x]/(x^2+x+1)$, so ${\mathbf Z}[(1+\sqrt{-3})/2]/P' = {\mathbf Z}[(1+\sqrt{-3})/2]/(2)$ is isom. to $({\mathbf Z}/2{\mathbf Z})[x]/(x^2+x+1)$, which is a field of size 4.)