= Solution
<Prime elements> are first-order definable in the <ring> <first-order language>. For instance let $\operatorname{Unit}(u)$ mean $\exists v\ (uv=1)$ and use
$$
\operatorname{Prime}(x)\equiv x\ne0\wedge\neg\operatorname{Unit}(x)\wedge
\forall u\forall v\ (uv=x\to\operatorname{Unit}(u)\vee\operatorname{Unit}(v)).
$$
In the <integer> <ring> this defines <prime elements> up to the harmless choice of sign. Expand the <first-order language> by names for every <integer> and a new constant $c$. Take the <elementary diagram> of $\mathcal Z$ together with
$$
\operatorname{Prime}(c),\qquad c\ne\bar a\quad(a\in\mathbb Z).
$$
Every finite <subset> is <satisfiable> in $\mathcal Z$: choose a positive prime <integer> outside the finitely many excluded values, using the <Euclid's theorem>. <Compactness theorem> gives a <first-order model> of the entire <set>. The complete-diagram argument embeds $\mathcal Z$ elementarily into its reduct; identify that copy with the original <integer> <ring>.
The resulting <elementary extension> contains $c$, a <prime element> unequal to every named standard <integer>. Therefore
$$
\boxed{\mathcal Z\preccurlyeq\mathcal Z^*,\qquad
\mathcal Z^*\models\operatorname{Prime}(c),\qquad c\notin\mathbb Z.}
$$
The construction rules out all standard <integers>, not merely the finite list appearing in any one approximation.
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