Ideal generated by 3 and $1+\sqrt{-5}$ is not a principal ideal in the ring $\Bbb Z[\sqrt{-5}]$

It’s a theorem in algebraic number theory that if $R$ is the ring of integers of a number field $K$ and $z\in R$, then the field-theoretic norm of $z$ is equal to the cardinality of $R/(z)$. Here, $K=\mathbb Q(\sqrt{-5}\,)$ and $R$ is the $\mathbb Z$-span of $\{1,\sqrt{-5}\,\}$, i.e. $R=\mathbb Z[\sqrt{-5}\,]$. Since $R/(3)$ has cardinality $9$, the larger ideal $(3,1+\sqrt{-5}\,)=I$ has the property that $R/I$ has cardinality $3$. But the norm of $a+b\sqrt{-5}$ is $a^2+5b^2$. This never takes the value $3$, that is, there is no $z$ of norm $3$, so the ideal is not principal.

A usual procedure to treat this kind of questions is to consider a "norm" on these rings. For $z=a+b\sqrt{-5}$ with $a,b\in\mathbb Z$ define $N(z)=z(a-b\sqrt{-5})=a^2+5b^2\in\mathbb Z$. Then $N(zw)=N(z)N(w)$ for all $z,w\in\mathbb Z[\sqrt{-5}]$, which implies in particular that $z$ is a unit in $\mathbb Z[\sqrt{-5}]$ iff $N(z)=1$.

Suppose that $I=\langle3,1+\sqrt{-5}\rangle$, the ideal in $\mathbb Z[\sqrt{-5}]$ generated by $3$ and $1+\sqrt{-5}$, is principal, say $I=\langle z\rangle$, with $z=a+b\sqrt{-5}$ and $a,b\in\mathbb Z$. The elements $3$ and $1+\sqrt{-5}$ obviously belong to $I$, so they are multiples in $\boldsymbol{\pmb{\mathbb Z}[\sqrt{-5}]}$ of the generator $z$. Therefore we have that $N(z)$ divides both $N(3)=9$ and $N(1+\sqrt{-5})=6$ in $\pmb{\mathbb Z}$, and so $N(z)$ divides $9-6=3$. In other words, $a^2+5b^2=1$ or $3$, so necessarily $b=0$ (otherwise $a^2+5b^2\geq5$), which in turn implies $a^2=1$ (because $3$ is not a square in $\mathbb Z$), so $z=\pm1$, and in particular $\pm z\in I$, so $1\in I$.

Thus, we can write $1=3(c+d\sqrt{-5})+(1+\sqrt{-5})(e+f\sqrt{-5})$, with $c,d,e,f\in\mathbb Z$, which implies $1=3c+e-5f$ and $0=3d+e+f$. Taking classes modulo $3$ we obtain $1=e+f$ and $0=e+f$, which is impossible. This contradiction shows the result.