Yes, because (quadratic) number rings are easily shown to have dimension at most one (i.e. every nonzero prime ideal is maximal). But $\rm PID$s are precisely the $\rm UFD$s which have dimension $\le 1.\, $ Below is a sketch of a proof of this and closely related results.
Theorem $\rm\ \ TFAE\ $ for a $\rm UFD\ D$
$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le 1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1,\, $ i.e. $ $ coprime $\Rightarrow$ comaximal
$(5)\ \ $ $\rm D$ is Bezout, i.e. all ideals $\,\rm (a,b)\,$ are principal.
$(6)\ \ $ $\rm D$ is a $\rm PID$
Proof $\ $ (sketch of $\,1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$
$(1\Rightarrow 2)$ $\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$
$(2\Rightarrow 3)$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)$ $\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)$ $\ $ Ideals $\neq 0\,$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max
Remark $ $ Examples of non-PID UFDs are easy in polynomial rings: if $D$ is a non-field domain then it has a nonzero nonunit $d$ so by here the ideal $(d,x)$ is not principal.
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.)
Best Answer
$\mathbb{Z}[\sqrt{-5}]$ can be written as $\mathbb{Z}[x]/(x^2+5)$.
Hilbert's basis theorem says that, if a ring $R$ is noetherian, then so is $R[x]$. Also, if $I$ is an ideal of a noetherian ring $R$, then $R/I$ is also noetherian (this can be seen by looking at the generators of the inverse image of an ideal of $R/I$ under the standard projection map).
Since $\mathbb{Z}$ is noetherian, these two facts then imply that $\mathbb{Z}[x]/(x^2+5) \cong \mathbb{Z}[\sqrt{-5}]$ is noetherian, and so, in particular, satisfies the ACC on principal ideals.