Assume we have a countable, non-standard model of Peano Arithmetic (PA) in ZFC. http://en.wikipedia.org/wiki/Non-standard_model_of_arithmetic
Let $N^*$ be the universe of this model and let $m \in N^*$ be a non-standard natural number larger than any standard natural number. $2^m \in N^*$ is the size of the $P(m)$, the powerset of numbers less than m, inside the model. We can even encode a bijection between $P(m)$ and $2^m$ inside the model. For example, we can encode a set as an m-bit string. x is a member of the set if the x'th bit of the string is true. Each such string is the binary representation of a natural number less than $2^m$.
Skolem's paradox considers a set that is uncountable inside the model yet countable outside the model. Here, we have a finite bijection inside the model which seems to be uncountable outside the model.
My question is how can the countably infinite set $X = \{x : x \in N^* \land x < m\}$ have a countable powerset? How can $Y = \{y : y \in N^* \land y < 2^m\}$ be a countable set inside ZFC? We can show there is a bijection between $P(X)$ and $Y$.
I know the answer must be something like the model doesn't "see" all of $P(m)$. If this is the answer, would someone give an example of a set of standard natural numbers the model can't "see"? For example, does there exist a binary representation of a standard real number such that this binary string is not a proper subset of the binary representation of some non-standard natural number inside the model.
First of all, models of $\sf PA$ don't see sets, they see integers. Perhaps what you wanted to say that you have $N$ which is a non-standard model of $\sf ZFC$ whose integers are a non-standard model of $\sf PA$.
Other than that, the reason that you give is correct. Yes, there are uncountably many subsets to that [proper] initial segment of $\omega^N$, but $N$ is countable and don't know them all, and moreover it thinks that collection is finite. An example of a standard set of integers not in the model is easy: $\omega$ itself. If it were there, it would have to be equal to $\omega^N$ in which case the integers were standard.