$K$-valued functions on an infinite set

As discussed in the question you linked, maximal ideals of $K^I$ correspond to ultrafilters of $I$, and consequently, rings of the form $A/\mathfrak{m}$ are the same as ultrapowers of $K$. With that in mind, we can approach an answer to your first question.

A positive answer to your first question is not provable from ZFC (assuming ZFC is consistent). A well known example of the fields you discuss is a field of hyperreal numbers which is any ring of the form $\mathbb{R}^\mathbb{N}/\mathfrak{m}$ for a maximal ideal $\mathfrak{m}$ of $\mathbb{R}^\mathbb{N}$ containing $\oplus_{n=0}^\infty\mathbb{R}$. As stated by wikipedia and implied by Theorem 1 of this paper, if CH fails, then there will exist non-isomorphic ultrapowers of $\mathbb{R}$. In fact, $\mathbb{R}$ can be replaced with any real closed field, and the same incident will occur.

However, it gets worse. If $K$ is infinite and the ideal $\mathfrak{m}$ corresponds to what is called a $\textit{regular}$ ultrafilter of $I$, then the cardinality of $A/\mathfrak{m}$ will be the same as that of $A$ due to a theorem attributed to Keisler. Yet, there may exist nonregular ultrafilters. Indeed, Magidor shows that if the existence of a huge cardinal is consistent with ZFC, then it is consistent with ZFC that there exists a set $I$ and ultrafilter $U$ on $I$ such that the ultraproduct $\omega_0^I/U$ has a strictly smaller cardinality than $\omega_0^I$ does. As such, if we take $K$ to be any countably infinite field, then with the assumptions just stated, there could exist maximal ideals $\mathfrak{m}$ and $\mathfrak{n}$ for which $A/\mathfrak{m}$ and $A/\mathfrak{n}$ do not even have the same cardinality. (Edit: This is overkill. We can obtain a nonregular free ultrafilter by just choosing one which is not not uniform i.e. containing a subset of $I$ with cardinality less than that of $I$. Consequently, this shows in ZFC alone that $A/\mathfrak{m}$ and $A/\mathfrak{n}$ can have different cardinalities. Thanks for the catch Eric Wofsey.)

Edit 1: Due to a theorem of Chang and Keisler in their book "Model Theory", if $I$ is countable, $K$ is a field of cardinality at most $2^{\aleph_0}$, and CH holds, then all ultrapowers of $K$ are isomorphic. However, this does not answer the question when $K$ has cardinality greater than $2^{\aleph_0}$ or $I$ is uncountable and CH holds, so I would consider Pierre's first question still not entirely solved. As I noted earlier, the answer to Pierre's first question is negative if CH fails, but I would be interested to know if the question is provable or refutable (or neither) in ZFC+CH.

Edit 2: I will expand on Tim Campion's Math Overflow comment which yields a negative answer to the second question. It is a consequence of Los's theorem that any structure is elementarily equivalent to all of its ultrapowers, meaning that they satisfy the same first order sentences. In particular, for each positive integer $n$, the sentence $$\forall a_0\forall a_1\ldots \forall a_n\exists x(a_nx^n+\cdots+a_1x+a_0=0)$$ holds in $K$ if and only if it holds in $A/\mathfrak{m}$. Hence, if we choose $K$ to be an algebraically closed field, then $A/\mathfrak{m}$ will be as well. However, if $A/\mathfrak{m}$ were purely transcendental over $K$, we would have $A/\mathfrak{m}=K(S)$ for some algebraically independent set $S\subset A/\mathfrak{m}$. But $K(S)$ is not algebraically closed when $S$ is nonempty since the polynomial $x^2-s$ is irreducible over $K(S)$ for $s\in S$. If $S$ is empty, then $A/\mathfrak{m}=K$ which need not be the case.