Apparently, we can use compactness theorem to construct an infinitely large natural number such that it is not divisible by any (standard) natural number $n \in \mathbb{N}_{>1}$. And I must say that I have no idea how to do this.
I have seen the construction of a model of the theory of $\mathbb{N}_0$ containing an infinitely large natural number. The method is very similar to the one that is described in https://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic. It states that a new constant $c$ is added in a set of axioms $P*$, which is defined in a language including the language of Peano arithmetic.
So I thought that I should use a similar method to construct an infinitely large natural number such that it is not divisible by any (standard) natural number $n \in \mathbb{N}_{>1}$. The problem is what new constant should I add?. What's the idea to find this new constant? Furthermore, let's say that this new constant is $x$. Should I just infinitely many new axioms $(n < x)$?
Fix $\mathcal{L} = \{+,\times,<,(c_i)_{i \in \mathbb{N}}\}$ and consider the theory of the $\mathcal{L}$-structure $M=(\mathbb{N},+,\times,<,(i)_{i \in \mathbb{N}})$. Now take the language $\mathcal{L}'=\mathcal{L} \cup \{c^*\}$ where $c^*$ is a constant symbol, and the theory given by $Th_{\mathcal{L}}(M) \cup \Psi$ where $\Psi$ consists of the following $\mathcal{L}'$-sentences:
Now show that every finite subset of $Th_{\mathcal{L}}(M) \cup \Psi$ has a model (hint: you can take $M$ and choose a suitable interpretation of $c^*$). By compactness you get that $Th_{\mathcal{L}}(M) \cup \Psi$ is consistent. So take a model $M^*$ of $Th_{\mathcal{L}}(M) \cup \Psi$. You can check that $M \prec M^*$ (after taking the $\mathcal{L}$-reduct of $M^*$) and therefore the interpretation of $c^*$ in $M^*$ can be thought of as a non-standard natural number with the desired properties.
When using compactness arguments you can only expect to show that something exists, you usually won't be able to hold it in your hands and 'interpret'.