Noetherian ring with finitely many height $n$ primes

This is true - in a Noetherian ring, the only possible heights for which there are finitely many primes of that height are either $0$, or maximal ideals. This follows from the following fact:

Proposition: If $R$ is Noetherian and $\mathfrak p_0 \subsetneq \mathfrak p_1 \subsetneq \mathfrak p_2$ is a chain of distinct prime ideals in $R$, then there are infinitely many primes $\mathfrak q$ such that $\mathfrak p_0 \subsetneq \mathfrak q \subsetneq \mathfrak p_2$.

Proof: Localizing at $\mathfrak p_2$ and quotienting by $\mathfrak p_0$, it suffices to show that any Noetherian local domain of dimension $2$ has infinitely many primes. If there were only finitely many height $1$ primes, then since $\mathfrak p_2$ is not contained in any of them, by prime avoidance it is not contained in their union, so there exists $x \in \mathfrak p_2$, but not in any height $1$ prime. But this contradicts the Principal Ideal Theorem, since $\text{ht}(x) \le 1$.

Suppose that $0<n<d=\dim R$. Let $\mathfrak p$ be a prime ideal of height $d$. There is a chain of prime ideals $\mathfrak p_0\subset\cdots\subset\mathfrak p_d=\mathfrak p$. Somewhere in this chain there is a prime ideal $\mathfrak q$ of height $n$ which is contained in a prime ideal $\mathfrak q'$ of height $n+1$ and contains a prime ideal $\mathfrak q''$ of height $n-1$. By Theorem 144 from Kaplansky, Commutative Rings, it follows that there are infinitely many prime ideals between $\mathfrak q''$ and $\mathfrak q'$ and all have height $n$.