Existence of nonstandard elementary extensions of $PA$?

Question

My question follows from the 1958 result of MacDowell–Specker (located originally in Modelle der Arithmetik, J. Symbolic Logic Volume 38, Issue 4 (1973), 651-652) of the proof of the following theorem:

Theorem 1. Every model $M$ of $PA$ has a proper elementary end extension $E_M\subset M$, where we take $\subset$ to mean $E_M\subseteq M$ and $\neg (E_M=M$).

The question I ask is broken into two parts which I believe are answerable in one post. Intuitively it feels that they are related as questions by the Macdowell-Specker as stated below:

Theorem 2. Every nonstandard model $N$ of $PA$ has a proper elementary end extension with $E_N\subseteq N$.

Question 1) Following from the existence of the proper end extension $E_M$ by the Macdowell-Specker theorem, can one assume that $E_N\models\mathrm{PA}$ if each $E_i\subset N$ is nonstandard? Perhaps more briefly, I ask if $E_N\models\mathrm{PA}$ still holds when $E_N$ is not a proper elementary extension of $N$.

Question 2) By the assumption of the "nonstandardess" of $E$, is it a sufficient condition for [Theorem 2] that if $E_N\models\mathrm{PA}$ then $N$ has its proper elementary end extensions $|E_N|\leq \omega$?

Thank you for your help, and I apologize if only one question should be asked in this context. The only similar question on the website I can find possibly related to the above is https://mathoverflow.net/questions/141845/elementary-end-extensions-of-models-of-peano-arithmetic-in-uncountable-languages, but this question concerns uncountability, which I do not feel is straightforwardly related to my question.

Answer 1

I think I understand the question well enough to answer now.

1. Every model is a (proper, by definition) elementary end-extension of itself, so it's only interesting to prove that there are also elementary end-extensions which aren't proper.
2. It follows easily from the ordinary proof of Macdowell-Specker that when $M$ is a countable model of PA, $M$ has a countable proper elementary end-extensions.
3. It follows that if $M$ is countable, there must also be uncountable elementary end-extensions: let $M=M_0$, given $M_\alpha$, let $M_{\alpha+1}$ be a proper elementary end-extension of $M_\alpha$, and when $\lambda$ is a limit ordinal, le $M_\lambda=\bigcup_{\alpha<\lambda}M_\alpha$. Then $M_{\omega_1}$ is an uncountable elementary end-extension of $M$.