I want to prove the statement in the title using Godel's theorems. Without Godel's theorem, I may use the theorem about infinite model from cardinality $\kappa$ $\Rightarrow$ infinite models from any cardinality $>\kappa$ to get as many non-isomorphic models as I wish (cause $\mathbb{N} \models \Sigma$)
I was trying to use Godel's theorem in the next manner, every recursive $\Sigma \subset Th(\mathbb{N})$ is not complete (and so for finite subsets as they are recursive), therefore one can find sentence $\alpha$ such that $\Sigma\nvdash \alpha , \neg\alpha$ so $\Sigma\cup\alpha$ , $\Sigma\cup\neg\alpha$ has models $M_1 , M_2$ because there are consistent ($\Sigma\subset Th(\mathbb{N})$ and therefore consistent ) . $M_1,M_2$ are not isomorphic cause they are not elementary equivalent.
I don't find an Idea to generalize this approach to get $2^{\aleph_0}$ models, any other approach will be appreciated, as long as it is using Godel's theorems
You can use this idea to build a whole infinite binary tree of theories.
At each node in the tree you have a finite set of sentences $B$ such that $PA+\Sigma+B$ is consistent. Apply the Gödel-Rosser construction to this theory to get an undetermined $\alpha$. The two successors of the node will arise by appending either $\alpha$ or $\neg\alpha$ to $B$.
The tree has countably many nodes, but $2^{\aleph_0}$ many infinite branches. Each branch has a model thanks to compactness. The models of two different infinite branches are non-isomorphic because they disagree about the $\alpha$ where their branches diverge from each other.