Let $R$ be a commutative Noetherian ring having the property that for every $R$-module $M$ that has finite projective dimension, every submodule of $M$ also has finite projective dimension. Then $R$ is obviously regular since every ideal has finite projective dimension.
My question is: Does $R$ have finite global dimension ?
Yes.
If $R$ has infinite global dimension, then it has a module $X$ of infinite projective dimension. But if $P_0\to X$ is a surjective map from a projective module to $X$, with kernel $Y$, then $Y$ must have infinite projective dimension, since if $Y$ had a finite projective resoltion $$\cdots\to 0\to P_n\to\cdots\to P_2\to P_1\to Y\to0$$ then $X$ would have a finite projective resolution $$\cdots\to 0\to P_n\to\cdots \to P_1\to P_0\to X\to0.$$
So $Y$ is a module of infinite projective dimension that is a submodule of $P_0$, which is projective and so certainly has finite projective dimension.