I am trying to proof that each element of $R :=\mathbb{Z} (\sqrt{-5})$ is factorable into irreducible elements.
Let $x \in R, x = AB$ with $A = (a + b \sqrt{-5} ),$ $B = (c + d \sqrt{-5})$.
If $A = \pm 1$ or $B = \pm 1$ then $x$ irreducible and the proof is done.
Otherwise, $x = AB, N(x) = N(A) N(B)$ with $N(A),N(B) \neq 1$.
How can I proceed?
Thanks in advance!
Your idea in the comment is right (see the third proof below).
An nonzero and noninvertible element $a$ of a domain $R$ fails to be a product of irreducible elements if, for every $b_1,b_2,\dots,b_n\in R$, with $b_i$ not invertible, at least one of the factors $b_i$ can be written as $b_i=cd$, where neither $c$ nor $d$ is invertible, that is, is not irreducible.
Assume $a$ is not a product of irreducibles. In what follows all elements I write as $a_{ij}$ are assumed to be noninvertible. Since $a=a_{11}$ is not irreducible, $a_{11}=a_{21}a_{22}$ and it's not restrictive to assume $a_{22}$ not irreducible, so we can factor it and get $$ a_{11}=a_{21}a_{22}=a_{31}a_{32}a_{33} $$ where the elements are reordered so that $a_{33}$ is not irreducible. $$ a=a_{n1}a_{n2}\dotsm a_{nn} \tag{*} $$ This allows us to build the increasing chain of ideals $$ (a)=(a_{11})\subsetneq(a_{2i_2})\subsetneq\dots\subsetneq(a_{ni_n}) \subsetneq\dotsb $$ where at each stage we choose one of the terms we factored, and therefore $R$ is not Noetherian.
The ring $\mathbb{Z}[\sqrt{-5}]$ is a homomorphic image of $\mathbb{Z}[X]$ via the isomorphism $\mathbb{Z}[\sqrt{-5}]\cong\mathbb{Z}[X]/(X^2+5)$, so it is Noetherian, because $\mathbb{Z}[X]$ is Noetherian by Hilbert’s Basis theorem.
However, the specific case is easier. From (*), applying the norm function $$ N(x+y\sqrt{-5})=x^2+5y^2 $$ we get $$ N(a)=N(a_{n1})N(a_{n2})\dotsm N(a_{nn}) $$ where $N(a_{nj})>1$, for $j=1,2,\dots,n$. Since $N$ has (positive) integer values, this is a contradiction, because it would follow that $$ N(a)\ge2^n $$ for every $n\ge1$.
Alternatively, do an induction on $N(a)$. If $N(a)=2$, the statement is (vacuously) true. Suppose it holds for every element $b$ with $2\le N(b)<N(a)$. Then either $a$ is irreducible and we are done, or $a=b_1b_2$, with $N(b_1)>1$ and $N(b_2)>1$; then $2\le N(b_1)<N(a)$ and $2\le N(b_2)<N(a)$ and the induction hypothesis applies.
Why showing the big gun proof? Because the same result in every Noetherian domain, not just a simple integral extension of the integers.
Note however that uniqueness of factorization need not hold. In particular it doesn't hold in $\mathbb{Z}[\sqrt{-5}]$, because $$ 6=2\cdot 3=(1+\sqrt{-5})(1+\sqrt{-5}) $$ where $2$ and $3$ are associate neither of $1+\sqrt{-5}$ nor of $1+\sqrt{-5}$.
