Sufficient conditions for being a PID

Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, can we conclude that R is a PID?

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 $(\Leftarrow)$ note that ACCP implies that the divisibility relation is well-founded, hence a nonzero ideal I 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.

No, in general this is not true. For an example, in the ring of entire functions every finitely generated ideal is principal, but this ring is not a PID.

If, in addition, $R$ satisfies the ascending chain condition for principal ideals, then one can conclude that $R$ is a PID.