A domain with every two-generated ideal principal $\rm\:(a,b) = (c)\:$ is called a Bezout domain. By induction this is equivalent to every finitely-generated ideal being principal. This does not imply that every ideal is principal (PID). Indeed, the ring of all algebraic integers is Bezout (Kaplansky, Commutative Rings, Theorem 102), but its ideal $\:(2^{1/2},\,2^{1/3},\ldots)\:$ is not principal.
A domain is a PID iff it's Bezout and satisfies ACCP (ascending chain condition on principal ideals). The direction $(\Rightarrow)$ is clear. For $(\Leftarrow)$ Hint: $ $ ACCP $\Rightarrow $ among $\rm(i)\subseteq I$ there is a maximal one $\rm(c),$ so $\rm \,(c,i) = (c)\,\ \forall\,i\in I,\,$ so $\,\rm I=(c)$. Or, said element-wise: ACCP implies divisibility is well-founded, so an ideal $\rm\,I\neq 0\,$ is generated by an element c minimal wrt to divisibility, since if $\rm\:c\nmid i\in I\:$ then $\rm\:(c,i) = (d)\subset I\:$ and $\rm\:d\:|\:c\:$ properly (else $\rm\,c\mid d\mid i\,),$ contra minimality of c. This is essentially an application of the Dedekind-Hasse Test for a PID.
If we are given a Bezout UFD $(\Rightarrow$ ACCP) then we can use the number of prime factors as a metric for the Dedekind-Hasse test. Then that the above $\rm\,c\in I\,$ is divisibility minimal translates to it being an element of $\,\rm I\,$ with least number of prime factors.
Best Answer
Answer: no. $\mathbb{Q}[X]$ is a PID, whereas $\mathbb{Z}[X]$ is not. See this