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,
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  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.
 Integral extensions and integral dependence: pdf file --- html page.