You can find encodings in the natural numbers of certain collections of subsets of $\mathbb{N}$. For example, as was pointed out by Henning Makholm, there is a very nice encoding of the collection of finite subsets of $\mathbb{N}$.
More generally, let $A$ be a recursively enumerable subset of $\mathbb{N}$. Let $T_A$ be a Turing machine that, on input $n$, halts if $n\in A$, and does not halt otherwise. Then we can encode $A$ by using the usual index of the machine $T_A$.
This encoding is not fully satisfactory. There are infinitely many Turing machines that halt on input $n$ iff $n\in A$. Thus every recursively enumerable set has infinitely many encodings. Moreover, there is no algorithm for telling, given two natural numbers, whether they encode the same set.
For more modest collections than the collection of recursively enumerable sets, there are far more satisfactory encodings. As a small and not very interesting example, consider the collection of subsets of $\mathbb{N}$ that are either finite or cofinite (their complement is finite). A small modification of the encoding of finite subsets will take care of the finites and cofinites. Basically, just use the even natural numbers for the finites. If you have used $2k$ to encode a finite, use $2k+1$ to encode its complement.
In addition, the elements of any countably infinite subset of the power set of $\mathbb{N}$ can, by the definition of countably infinite, be encoded using the natural numbers. However, by cardinality considerations, we cannot encode all the subsets of $\mathbb{N}$ using elements of $\mathbb{N}$.
Note that if $A\in\bigcup_{n=1}^\infty\mathcal P(C_n)$, then there exists some $n$ such that $A\in\mathcal P(C_n)$. Therefore $A\subseteq\{1,\ldots,n\}$ and so $A$ must be finite.
So indeed $\Bbb N$ itself is not in this union, despite the fact that the union is infinite, since $\Bbb N$ is not a finite set for itself.
Best Answer
Yes. For every $n \in \mathbb{N}$, the set $\mathbb{N} \setminus \{ n \}$ is in $P(\mathbb{N})$, and is infinite. And, there are clearly as many of those as there are natural numbers, i.e. infinitely many.