I know that every non-zero prime ideal in a Dedekind Domain is maximal. Since $\mathbb Z[\sqrt 5]$ is not integrally closed in its field of fractions $\mathbb Q[\sqrt 5]$, I wonder whether we can still show that non-zero primes are maximal.
More generally is it known for which of the square-free $D\in \mathbb Z$ there are non-zero primes of $\mathbb Z[\sqrt D]$ that are not maximal.
The ring $\mathbb{Z}[\sqrt{D}]$ is integral over $\mathbb{Z}$. Therefore a prime ideal $P$ of it is maximal if and only if the prime ideal $p:=P\cap\mathbb{Z}$ is maximal. Since all non-zero prime ideals of $\mathbb{Z}$ are maximal we get that all $P$ such that $P\cap\mathbb{Z}\neq 0$ are maximal. The latter condition is equivalent to $P\neq 0$, which proves the claim.
Below is a very simple direct proof for any ring $\,R\,$ of algebraic integers.
Lemma $\ $ If $\ I \supsetneq P\ $ are ideals of $\,R,\,$ with $\,P\,$ prime then there is an integer $\,f_k\in I\,$ but $\,f_k\not\in P\,.$
Proof $\ $ Choose $\,\alpha\in I,\ \alpha\not\in P\,.\,$ Being an algebraic integer, $\,f(\alpha) = 0\,$ for a monic $\,f(x)\in \mathbb Z[x],\ $ $\,f(x) \, =\, x^n +\cdots + f_1\ x +\, f_0\,.\,$ Note $\,f_n = 1\not\in P\,.\,$ Let $\,k\,$ be least with $\,f_k\,\not\in P.\,$ $\ f(\alpha) = 0\in P\,$ $\, \Rightarrow\,$ $\,(\alpha^{n-k}+\cdots+f_k)\ \alpha^k\in P,\ \alpha\not\in P$ $\,\Rightarrow\,$ $\,\alpha^{n-k}+\cdots+f_k\in P\subset I\,.\,$ So $\ \alpha\in I$ $\,\Rightarrow\,$ $\,f_k \in I\,.\ $ QED
Corollary $\, $ A proper chain of prime ideals in $\,R\,$ cannot contract to a shorter such chain in $\,\mathbb Z\,.$
Remark $\ $ Alternatively, reduce to the simpler case $\,P = 0\,$ by way of factoring out the prime $\,P\,.\,$ Then $\,f_k\,$ is the constant term of a minimal polynomial for $\,\alpha\,$ over a domain, so $\,f_k\ne 0,\,$ i.e. $\,f_k\not\in P,\,$ which is explained in much detail in this post, as a generalization of rationalizing denominators.
More generally, integral ring extensions preserve (Krull) dimension. Even more generally, it is preserved by ring extensions satisfyng both GU (going-up) and INC (incomparable). For a proof see Kaplansky, Commutative Rings, Theorem 48. For much more on precisely how GU,INC and LO (lying over) are related to integrality for ring extensions (and homs) see the paper by Dobbs cited in this answer.
After having read Bill Dubuque's beautiful answer, I'm rewriting mine.
To answer the question as asked, the simplest seems to prove this.
If $P$ and $I$ are ideals of $\mathbb Z[\sqrt5]$, and if $P$ is prime, then $$ P < I\quad\implies\quad\mathbb Z\cap P\ < \ \mathbb Z\cap I, $$ where $ < $ means "properly contained in".
To see this, set $x=a+b\sqrt5$ with $a,b\in\mathbb Z$, and let $x$ be in $I$ but not in $P$. Putting
$$
x':=a-b\sqrt5,\quad t:=x+x'=2a,\quad n:=xx'=a^2+5b^2,
$$
we have
$$
x^2-tx+n=0.
$$
If $n$ is not in $P$, we are done because $n$ is in $\mathbb Z\cap I$. If $n$ is in $P$, then so is $x-t$, and $t$ is in $\mathbb Z\cap I$ but not in $P$.
More generally, if $B$ is a (commutative) ring, if $A$ is a subring, if $B$ is integral over $A$, if $P$ and $I$ are ideals of $B$, and if $P$ is prime, then $$ P < I\quad\implies\quad A\cap P\ < \ A\cap I. $$ Indeed, on taking the quotient by $P$, we can assume that $B$ is a domain, and that $P=0$. Let $x$ be a nonzero element of $I$, and let $f\in A[X]$ a monic polynomial of least degree such that $f(x)=0$. Then the constant term of $f$ is a nonzero element of $I\cap A$.
In view of the second Corollary on page 7 of the text [1] by Mel Hochster, if a ring $B$ is integral over its subring $A$, then $A$ and $B$ have the same Krull dimension.
This shows in particular that if a ring $A$ is between $\mathbb Z$ and the ring of integers of a number field, then $A$ has Krull dimension $1$, that is, its nonzero primes are maximal.
[1] Integral extensions and integral dependence: pdf file --- html page.
