As mentioned, there are easy counterxamples. However, it is true for UFDs since PIDs are *precisely* the $\rm UFDs$ of dimension $\le 1,\:$ i.e. such that prime ideals $\ne 0$ are maximal. 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

$(6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$