As requested this is a slightly more detailed answer to your first question. I really don't know how to approach your second question.

By a mild variation Vaught's two cardinal to get a model $N$ of $PA$ with a non-standard number $n$ such that $|\{x\in N:x \leq n \}|=\aleph_0$ but $|\{x\in N : x\leq 2^n\}|=\aleph_1$ it's enough to find an elementary pair of models $M \prec N$ and a non-standard number $n\in M$ such that $\{x\in M : x\leq n\}=\{x\in N : x\leq n\}$ but such that $\{x\in M: x \leq 2^n \}$ is a proper subset of $\{x \in N : x\leq 2^n \}$.

Choose a constant $n$ and add it to the language of $PA$. Now assume for the sake of contradiction that in every model $N$ of $PA$ for any $n\in N$ if $\{x\in N:x \leq n \}$ is infinite, then $|\{x\in N:x \leq n \}|=|\{x\in N : x\leq 2^n\}|$. Then by theorem 12.1.5 in Hodges' big model theory textbook there must be a layering of the definable set $\{x\in N : x\leq 2^n\}$ by the definable set $\{x\in N:x \leq n \}$ which by exercise 7 in section 12.1 implies that there is some polynomial $p(x)$ with integer coefficients such that for any model $M$ of $PA$ (with the constant $n$) if $|\{x\in N:x \leq n \}|=m<\omega$, then $|\{x\in N : x\leq 2^n\}| \leq p(m)$, but this is clearly absurd since $2^n$ grows faster than any polynomial.

Therefore no such layering can exist and there must be a model $N$ of $PA$ with the constant $n$ such that $|\{x\in N:x \leq n \}|$ and $|\{x\in N : x\leq 2^n\}|$ are both infinite and $|\{x\in N:x \leq n \}| < |\{x\in N : x\leq 2^n\}|$. So by the Löwenheim–Skolem theorem there is an elementary substructure $M$ of $N$ such that $\{x\in N:x \leq n \} \subseteq M$ and $|\{x\in N:x \leq n \}|=|M|$, so that in particular $\{x\in N : x\leq 2^n\}$ is not a subset of $M$.

yes. $\endgroup$ – Noah Schweber Feb 15 at 1:19