As it is pointed out by Mohan in a comment, a commutative noetherian local ring is regular if and only if all modules have finite projective dimension.

As for the second question, you are right. Since $\sqrt{I}$ is the intersection of all the prime ideals containing $I$, if $\sqrt{I}=\mathfrak m$, then $\mathfrak m$ is the only prime ideal containing $I$. This means that in $R/I$ there is exactly one prime ideal, and then $\dim(R/I)=0$.