I can solve this question on the assumption that the $x_i$s are not zero-divisors since $\dim(R/(x)) = \dim(R)-1$ if $x$ is not a zero-divisor. My question is, how do I prove that they are not zero divisors?
I can't use the fact that Regular Local Rings are domains since that fact is proved just below, making use of this exercise. I have made the observation that the $x_i$s are a minimal set of generators for $\mathfrak{m}$, but I'm not sure if this leads to anything.
For regularity, you need to show that the longest chain of prime ideals in $R/(x_1,\ldots,x_i)$ has length no less than $d-i$. If you had a shorter chain of prime ideals, could you use some kind of correspondence lemma to lift it to a chain of prime ideals in $R$ and get a contradiction?
EDIT:
From Matsumura Commutative Ring Theory Theorem 13.6 you can deduce $\mathfrak m/(x_1,\ldots, x_i)$ has height $d-i$, which I think is all you need.