1) Is $e^{e^x}$ irrational for all rational $x$?
It is known that $e^x$ is transcendental for every nonzero algebraic $x$. But this dos not help here because for transcedental $x$, $e^x$ can be rational.
2) Is $e^{e^x}$ transcendental for all algebraic $x$?
This would imply 1).
I see you have also asked similar question here. It is a bit old, but nevertheless I believe I have results that might interest you, especially the later one.
Claim: $[2]$ is true, conditionally on Schanuel's conjecture
Proof: Specifically, Schanuel's conjecture states that
$$\text{trdeg}_{\mathbb{Q}} \,\! \mathbb{Q}\left(x_1, x_2, \cdots x_n, e^{x_1}, e^{x_2}, \cdots , e^{x_n}\right) \geq n$$
for $\mathbb{Q}$-linearly independent $x_1, x_2, \cdots, x_n \in \mathbb{C}$. One can verify that $(x_1 , x_2) = (e^x, 1)$ is $\mathbb{Q}$-linearly independent, thus the above gives $\left(e^x, e^{e^x}\right)$ is a $\mathbb{Q}$-algebraically independent pair and since $e^x$ for $x \in \mathbb{\bar Q}$ is transcendental, we have transcendence of $e^{e^x}$. $\blacksquare$
Claim: At least one of $\large e^{e^x}, e^{e^{2x}}$ and $\large e^{e^{3x}}$ is transcendental
Proof: This can be concluded from the lines of the proof of Schneider's conjecture done in Brownawell$(1974)$. In general, it has been proved (unconditionally, in the same paper) that for $\alpha, \beta \in \mathbb{C} \! \setminus \!\!\mathbb{Q}$ and $\gamma \in \mathbb{C}$ if $e^{\alpha}, e^{\alpha \gamma} \in \mathbb{\bar Q}$ then at least two of
$$\large \alpha, \beta, \gamma, e^{\beta\gamma}, e^{\alpha\beta\gamma}$$
are $\mathbb{Q}$-algebraically independent. The claim is then proved by setting $\alpha = \beta = \gamma = e^{x}$ $\blacksquare$
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