 Solution

The multiplicative field norm on this ring isIf an element is a unit, its positive integer norm has a multiplicative inverse in the positive integers, hence is one. The only integer solutions of $a^2+5b^2=1$ are $(a,b)=(1,0),(-1,0)$. Therefore the units are exactly $1$ and $-1$.

The standard failure of unique factorization isThere is no element of norm two or three: $b\ne 0$ gives norm at least five, while $b=0$ gives a square. An alleged nontrivial factorization of $2$ (norm four), $3$ (norm nine), or $1\pm \sqrt{-5}$ (norm six) would require a factor of norm two or three. Thus all four displayed factors are irreducible elements. No factor in the first factorization is associated to a factor in the second, since associates have the same norm and the norms are respectively four, nine and six. Hence $\mathbb{Z}[\sqrt{-5}]$ is not a unique factorization domain.