Height of a contraction of prime ideal 
Let $P\neq 0$ be a prime ideal of $\mathbb{Z}[X_1,\ldots,X_m]$ of height $n$, i.e. the longest chain of prime ideals contained in $P$ has length $n$, with $P \cap \mathbb{Z} = 0$. 
  I want to show that (possibly after relabeling $X_1,\ldots,X_m$) the ideal $P' = P\cap\mathbb{Z}[X_1,\ldots,X_{m-1}]$ has height $n-1$. 

Intuitively it seems to be clear that since $P'$ is contained in $P$, the maximal length of a chain of ideals contained in $P'$ must be one less. The problem is that $P'$ is not an ideal in $\mathbb{Z}[X_1,\ldots,X_m]$. I cannot find a nice argument why such a relabeling always exists. It seems to be a standard result.
 A: Let me give another answer without using Cohen-Macaulay rings. (But the catenarity I don't think can be avoided.) 
Let $Q\subset P$ be prime ideals in $\mathbb Z[X_1,\dots,X_m]$ with $P\cap\mathbb Z=(0)$. Then $Q\cap S=P\cap S=\emptyset$, where $S=\mathbb Z\setminus\{0\}$, and therefore $S^{-1}Q\subset S^{-1}P$ are prime ideals in $$S^{-1}(\mathbb Z[X_1,\dots,X_m])=\mathbb Q[X_1,\dots,X_m]$$ of the same height as $Q$, respectively $P$. Since $\mathbb Q[X_1,\dots,X_m]$ is a polynomial ring over a field it is catenary, and then $\operatorname{ht}P-\operatorname{ht}Q=\operatorname{ht}(S^{-1}P)-\operatorname{ht}(S^{-1}Q)=\operatorname{ht}(S^{-1}P/S^{-1}Q)=\operatorname{ht}(P/Q)$. In other words, the polynomial ring $\mathbb Z[X_1,\dots,X_m]$ is also catenary.   
Now set $\mathfrak p_i=P\cap\mathbb Z[X_1,\dots,X_{i-1},X_{i+1},\dots,X_m]$. I claim that there is $i\in\{1,\dots,m\}$ such that $P\ne\mathfrak p_i[X_i]$. (Here $\mathfrak p_i[X_i]$ denotes the extension of $\mathfrak p_i$ to $\mathbb Z[X_1,\dots,X_m]$.) Suppose the contrary. Then $P=\mathfrak p_i[X_i]$ for all $i$. We show that this leads to $P\cap\mathbb Z\ne(0)$. Since $P\ne(0)$ all prime ideals $\mathfrak p_i$ are also non-zero. Let $a_1\in\mathfrak p_1$, $a_1\ne0$. (Note that $a_1$ is a polynomial in $X_2,\dots, X_m$.) Since $a_1\in\mathfrak p_2[X_2]$ all coefficients of $a_1$ (written as a polynomial in $X_2$) belong to $\mathfrak p_2$. Thus we get a non-zero polynomial $a_2\in\mathfrak p_2$. (Note that $a_2$ is a polynomial in $X_3,\dots,X_m$.) Continuing this way we get a non-zero constant polynomial $a_m\in\mathfrak p_m$, so $P\cap\mathbb Z\ne(0)$, a contradiction. 
Suppose $P\ne\mathfrak p_m[X_m]$. Then $1=\operatorname{ht}(P/\mathfrak p_m[X_m])=\operatorname{ht}P-\operatorname{ht}(\mathfrak p_m[X_m])=\operatorname{ht}P-\operatorname{ht}\mathfrak p_m$, so $\operatorname{ht}\mathfrak p_m=n-1$. Now set $P'=\mathfrak p_m$, and you are done.
