Given the language of arithmetics $L=\{0, 1, +, \cdot\}$ one should prove that the class of all non-standard models is not axiomatizable.
So basically we have (for $M$ - standard model of arithmetics):
$$S= \{M':M\cong M'\} \subset Mod^{PA}(L)\,\,\,\,\,\,\,\,\,\,\, NS = Mod^{PA}(L)\setminus S.$$
It is rather easy to see that $S$ is not countably axiomatizable thanks to Löwenheim-Skolem theorem. Is this related to the impossibility of axiomatization of non-standard models?
Thanks in advance.
Let $C$ be the set of all sentences in the language $L$ of $PA$ that are true in the standard model $\mathbb{N}$ (your $M$). As the theory of a specific model, $C$ is complete: i.e., for any $\phi$, either $\phi \in C$ or $\lnot \phi \in C$. Now add to $L$ a constant $c$ and consider the set of sentences $I = \{c > 0, c > 1, c > 2,, \ldots\}$. Any finite subset $A$ of $C \cup I$ has a model, e.g., $\mathbb{N}$, taking $c = n + 1$ where $n$ is the largest natural number such that $c > n$ appears in $A$. So, by compactness, $C \cup I$ has a model $M'$ say which is necessarily non-standard, since its interpretation of $c$ must be greater than every standard natural number. As $C$ is complete, any sentence $\phi$ that does not contain $c$ and is true in $M'$ must also be true in $\mathbb{N}$ (since otherwise $\lnot\phi \in C$ implying that $M'$ cannot exist). Hence no set of axioms in the language of $PA$ can have $M'$ as a model but not $\mathbb{N}$.