Are there any interesting non standard models of $Q$? I've encountered a few of the models of Robinson arithmetic here a quick list:
$\mathbb{N}\cup {\infty}$ (used to show Robinson arithmetic has a non standard model)
$\mathbb{N}\cup \{a,b\}$ where the operations are defined so $a$ and $b$ do not commute.
$\mathbb{Z}[x]^+$ which is the set of all integer polynomials with positive leading coefficient. (In this model you can show Robinson Arithmetic cannot prove all elements are odd or even.)
All these models seem to be used only to show how weak Robinson Arithmetic is or rather what things it's unable to prove. Besides the natural numbers are there some models that have some importance beyond being used as a counter example? I am aware that any model of $Q$ must contain an isomorphic copy of the natural numbers. But for weak theories like the group axioms there are plenty of important "models" for example the symmetric group is studied for other reasons besides proving that the group axioms can't prove commutativity.
 A: While $\mathbb{N}\cup\{\infty\}$ is a bit boring, it's an initial segment of a much more interesting model of $\mathsf{Q}$: namely, cardinal arithmetic (up to whatever cutoff point you want). (Note that we take $S\kappa=\kappa+1=\kappa$, not $\kappa^+$, when $\kappa$ is infinite.) Of course this is a very boring arithmetic system, but it's definitely relevant outside of $\mathsf{Q}$-flavored topics.
Ordinal arithmetic is more interesting, but it doesn't quite get $\mathsf{Q}$: successor and multiplication play a bit messily with each other. If we want $\omega$ to be a successor we need to interpret $S\alpha$ as $1+\alpha$ rather than $\alpha+1$, and that breaks the multiplication axioms. Still, ordinal arithmetic does form a very natural model of an "alternative arithmetic," just not $\mathsf{Q}$ specifically.

And here's another one coming from set theory which is much less trivial (if less useful as well):
Work in $\mathsf{ZF}$ (= set theory without choice). A Dedekind-finite set is one which does not have a non-surjective self-injection; every finite set is Dedekind-finite, but the converse is not provable in $\mathsf{ZF}$ alone. It's easy to prove that the Dedekind-finite sets are closed under finite unions, finite Cartesian products, and adding or removing (if the set is nonempty) single elements.
Ignoring set/class issues for simplicity, this leads to the following:

The collection of Dedekind-finite cardinalities consistently forms a nonstandard model of $\mathsf{Q}$.

Specifically, the collection of Dedekind-finite cardinalities is always a model of $\mathsf{Q}$, and it is nonstandard iff there is an infinite Dedekind-finite set.
Interestingly, there is a lot of flexibility re: how good the arithmetic of Dedekind-finite sets can be. In every "easily-constructible" model of $\mathsf{ZF}$ with Dedekind-finite sets, there are incomparable Dedekind-finite cardinalities and so we get a model of $\mathsf{Q}$ with no good ordering (in contrast to e.g. $\mathbb{Z}[x]^+$). On the other hand, it turns out that $(i)$ we can also (nontrivially) get all Dedekind-finite cardinalities comparable and $(ii)$ whenever we do so, the resulting model of $\mathsf{Q}$ in fact satisfies all of true first-order arithmetic! See here; note that there is a weak large cardinal assumption used, but its necessity is open.

As a coda, let me push back against the claim that $\mathbb{Z}[x]^+$ is unnatural. In my opinion, it's extremely natural; it's equivalent to $\mathbb{Z}[x]$ in a very strong sense, namely bi-interpretability, in the same way that we can build the integers from the natural numbers.
The issue is that negative elements are convenient in algebra but somewhat inconvenient in logic (since "induction for $\mathbb{Z}$" is messier than "induction for $\mathbb{N}$"). But everything we do algebraically with $\mathbb{Z}[x]$ could be appropriately (if tediously) rephrased to live entirely in $\mathbb{Z}[x]^+$, and conversely we could have whipped up "$\mathsf{Q}$-for-$\mathbb{Z}$" instead of $\mathsf{Q}$ (and similarly "$\mathsf{PA}$-for-$\mathbb{Z}$" and so on) with the same effect. So I think this is a case where what we're really seeing is a less convenient presentation of a common and extremely useful notion.
