Let $R$ be a Cohen-Macaulay ring with a prime ideal $P$ and an element $x\notin P$ such that the ideal $\langle x,P\rangle$ generated by $x$ and $P$ is not the whole of $R$. If, moreover, $Q$ is a minimal prime over $\langle x,P\rangle$ I want to show that the height of $Q$ is exactly one more than that of $P$.
I think we should first consider a descending chain of prime ideals starting at $P$ and adjoin $x$ to each prime in the chain. Then...?
Since $Q$ is minimal over $P+(x)$ it follows that $Q/P$ is minimal over $\bar x$ in $\bar R=R/P$. Thus $\operatorname{ht}Q/P=1$.
Now localize at $Q$ and reduce the problem to the following:
For CM local rings we know that $\dim R=\operatorname{ht}\mathfrak a+\dim R/\mathfrak a$ for any proper ideal $\mathfrak a$. When $\mathfrak a=\mathfrak p$ we get $\operatorname{ht}\mathfrak m=\dim R=\operatorname{ht}\mathfrak p+\dim R/\mathfrak p=\operatorname{ht}\mathfrak p+\operatorname{ht}\mathfrak m/\mathfrak p=\operatorname{ht}\mathfrak p+1.$