Prime in $\mathbb Z [\sqrt{-5}]$ but not in $\mathbb Z [\sqrt{5}]$

Yes, $\sqrt{-5}$ is irreducible and prime in $\mathbb{Z}[\sqrt{-5}]$. If it wasn't, it would be possible to find numbers in the domain such that $ab = c \sqrt{-5}$ yet $\sqrt{-5}$ divides neither $a$ nor $b$. Since the norm is multiplicative, that would require $N(ab) = 5$ and so the only possibilities are $N(a) = 1$, $N(b) = 5$ or vice-versa.

For the second part of your question, by $\mathbb{Z}[\sqrt{5}]$ do you mean $\mathbb{Z}[\phi]$ where $$\phi = \frac{1 + \sqrt{5}}{2}?$$ I'll assume that you do. Since $x^2 \equiv \pm 2 \pmod 5$ is insoluble in integers, we can be assured that primes ending in $3$ or $7$ are at least irreducible in both domains (and of course prime in $\mathbb{Z}[\phi]$).

So what we're looking for is that $x^2 \equiv -5 \pmod p$ has no solutions but $x^2 \equiv 5 \pmod p$ does. I'd work out the Legendre symbol for you but I'm running late to dinner.

The condition for $p$ to split in a quadratic field of discriminant $D$ is that

$$\left(\frac{D}{p}\right)=+1$$

The discriminant of $\mathbb{Z}(\sqrt{-5})$ is $D=-20$, and of $\mathbb{Z}(\sqrt{5})$ is $D=5$, thus you seek a $p$ such that

$$\left(\frac{-20}{p}\right)=-1, \left(\frac{5}{p}\right)=+1$$

Now $$\left(\frac{5}{p}\right)=\left(\frac{p}{5}\right)$$ and this is $+1$ if and only $p\equiv \pm 1 (\mod 5)$.

In this case

$$\left(\frac{-20}{p}\right)=\left(\frac{-1}{p}\right)$$ so $p$ must also satisfy $$\left(\frac{-1}{p}\right)=-1$$ or that

$p\equiv 3 (\mod 4)$.

The smallest such prime is $p=11$.

And indeed,

$$11=(4+\sqrt{5})(4-\sqrt{5})$$

and on the other had it is easy to see that

$$a^2+5b^2=11$$ is impossible.

One can also see that $19$ is another such prime. Thus in fact the set of such primes are those of the forms $20n+11$ and $20n-1$.