For what $x$ is there some $k$ such that $x\uparrow\uparrow k$ is an integer?

The Gelfond-Schneider theorem implies that

• if $x \upuparrows 2 = x^x \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is transcendental, and
• if $x \upuparrows 3 = x^{x^x} \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ then $x$ is irrational.

But I don't think it's known whether or not $x \upuparrows 3 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is transcendental, or whether or not $x \upuparrows 4 \in \mathbb{Z}$ and $x \not \in \mathbb{Z}$ implies that $x$ is irrational.

It is at least easy to see the following: for fixed $k$ the function $f(x) = x \upuparrows k$ satisfies $f(1) = 1$, is strictly increasing on $[1, \infty)$, continuous, and goes to infinity, so takes on every real value in $[1, \infty)$ exactly once. So for every positive integer $n \in \mathbb{N}$ there is a unique positive real $x \in [1, \infty)$ such that $f(x) = n$, which we might call the $k^{th}$ power-tower-root of $n$. But I think very little is known about these reals. I would not expect them to be expressible in terms of other familiar constants but I expect this to be quite out of reach. Here's another example of "this is obviously true and nobody has the slightest clue how to prove it": nobody knows whether or not $\pi + e$ or $\pi e$ are irrational (although at least one of them must be transcendental)!