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.
$\mathbb{Z}[\sqrt{-5}]$ 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.
Best Answer
The main point here is that to divide is to contain, which means that if $a \mid b$ then $(a) \supseteq (b)$.
A descending chain of divisors $$\cdots \mid d_n \mid \cdots \mid d_1 $$ corresponds to an ascending chain of ideals $$\cdots \supseteq (d_n) \supseteq \cdots \supseteq (d_1) $$
Now, consider $D=\bigcup_i (d_i)$. Then $D$ is an ideal and so $D=(d)$ for some $d$.
This means that the chain stops at the ideal $(d_n)$ that contains $d$.
In turn, this implies that $d$ is irreducible and you can continue your argument.