Yes, because such 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 having 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
$2)\ $ prime ideals are principal
$3)\ $ maximal ideals are principal
$4)\ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$
$5)\ $ $\rm D$ is Bezout
$6)\ $ $\rm D$ is a $\rm PID$
Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)
$1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$
$2\Rightarrow 3)$ $\ \: $ Clear.
$3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$
$4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$
$5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$