Here's my attempt to rigorously show that it cannot generalize. I originally wrote this for my blog, using the following formulation of the puzzle:

- All islanders are immortal, except as explained below.
- All islanders know the color of every other islander's eyes, but not their own.
- If an islander has enough information to logically deduce the color of their
own eyes, they will perish at midnight of the following day, which will be
noticed by the other islanders.
- Any message washed ashore in a bottle can be trusted.

Now as it happens, at 12:01am on Day 0, a message washes ashore in a bottle,
and is read by all the islanders:

There is at least one islander with blue eyes.

What happens?

## Finite solution

The following solution works
for infinitely many islanders, as long as only finitely many have blue eyes. I go into more information on my blog post, but since you already know the blue eyes puzzle you can probably prove the following.

**Lemma 1** If an islander sees at least $N$ other islanders with blue
eyes, she knows on Day $0$ that she will not be able to deduce her eye
color before Day $N$.

**Lemma 2** If an islander sees $N$ other islanders with blue eyes who
are alive on Day $N$, then she can deduce her eye color by Day $N+1$.

**Theorem 3** If there are $N+1$ islanders with blue eyes, they will
perish at midnight on Day $N+1$, and all others will survive.

## The infinite case

So, we've handled the case where there are an arbitrary number of islanders
(even infinite!), but only finitely many islanders with blue eyes. So,
what if there are infinitely many islanders with blue eyes?

Well, we already have a very nice lemma above, **Lemma 1**. That allows us
to prove that, since all islanders see infinitely many pairs of blue eyes,
no islander can deduce their eye color by Day $N$ for all
$N\in\{0,1,2,\dots\}$.

So, assuming that our days are limited to the finite ordinals, the case is
closed. But why should we stop there? What if there exists a day for every
ordinal number?

We should consider what happens on Day $\omega$. Will anyone wake up dead?
The answer is of course, no. All islanders were able to deduce from the
beginning that no one could perish after a finite number of days, so
by the time Day $\omega$ begins, no one has gained information they didn't
already have. We shall be able to extend this argument by
transfinite induction to handle every ordinal
(regardless of the cardinality of the ordinal or the set of blue-eyed islanders).

**Theorem 4** If there infinitely many islanders with blue eyes, then no islanders
can deduce their eye color by Day $\alpha$, where $\alpha$ is any ordinal.

*Proof.* If $\alpha$ is finite, we are done by Lemma 1. By transfinite induction,
we may assume the theorem holds for all $\beta\lt\alpha$. In order to
deduce her eye color by Day $\alpha$, $Alice$ requires new information on
a previous day. However, by the induction hypothesis, $Alice$ knows from
the beginning that every islander cannot deduce their eye color by a previous
day. So, if $\alpha$ is a limit ordinal, we are done. The other case
is when $\alpha=\beta+1$ for some $\beta\lt\alpha$. Alice wakes up
on Day $\beta$, and is immediately bored. She already knew that everyone
sees infinitely many islanders with blue eyes, so she already knew that
everyone would be alive. Indeed, they are. So, she has gained no new information
that morning, and therefore cannot deduce anything new by the end of the day,
the beginning of Day $\alpha$. $\square$

otherislanders map naturally to an ordinal (the days), allowing the induction to proceed. But with a countably infinite population, there's nonaturalway for an islander to map the other islanders to a particular countable ordinal. You might be able to do something by putting the islanders into some kind of hierarchy, but then that breaks the symmetry that is such a pleasing feature of the original problem. $\endgroup$