= Solution
In a <principal ideal domain>, an ascending chain of <ideals> stabilizes: its union is an <ideal> $(d)$, and its generator belongs to one member, which then contains the union. This proves the needed <ascending chain condition>. If a nonzero nonunit had no finite irreducible factorization, split it into two nonunits and select a factor that still has no such factorization. Continuing would give a strictly ascending chain of principal <ideals>, because $a=bc$ with $c$ a nonunit implies $(a)\subsetneq(b)$. This contradicts the chain condition, so factorization exists.
An <irreducible element> $p$ generates a <maximal ideal>: any intermediate <ideal> is $(d)$ with $p=dc$, and irreducibility forces $d$ or $c$ to be a unit. Therefore $p$ is a <prime element>. In any two irreducible factorizations, the first prime factor divides a factor of the other, hence is associate to it. Cancel and continue. This proves \b[every <principal ideal domain> is a <unique factorization domain>].
In $\mathbb Z[\sqrt{-3}]$, the multiplicative <field norm> is $N(a+b\sqrt{-3})=a^2+3b^2$. The only <units> are $\pm1$. There is no element of norm two, so every element of norm four is irreducible: a factorization into nonunits would require norms two and two. Consequently
$$
\boxed{4=2\cdot2=(1+\sqrt{-3})(1-\sqrt{-3})}
$$
is a pair of genuinely different irreducible factorizations. Neither $1+\sqrt{-3}$ nor $1-\sqrt{-3}$ is associate to $2$.
Take the <Eisenstein integers> $R=\mathbb Z[\omega]$, where $\omega=(-1+\sqrt{-3})/2$ and $\omega^2+\omega+1=0$. Since $\sqrt{-3}=1+2\omega$, the embedded subring is $\mathbb Z+2\mathbb Z\omega$, whose additive index in $R$ is two. Its larger ring has <Eisenstein-integer norm> $N(a+b\omega)=a^2-ab+b^2$ and is Euclidean. Indeed, write a complex quotient in the real basis $1,\omega$ and round both coordinates to integers. The coordinate errors $u,v$ satisfy $|u|,|v|\le1/2$, giving $|u+v\omega|^2=u^2-uv+v^2\le3/4<1$. Thus division leaves a remainder of smaller norm. A <Euclidean domain> is a <principal ideal domain>, because division by a nonzero element of least norm in an ideal leaves zero remainder. This gives the requested \b[index-two embedding into a <principal ideal domain>].
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