Expanding on some of this...
I suppose it makes no sense unless $x$ is between 0 and 1. And in this case...
Define $y =x^{x^{x^{\cdots}}}$. Then we have $y= x^y$. For any fixed $0<x<1$, this defines a unique positive value for $y$ (intermediate value theorem will imply $y$ is between 0 and 1).
So that's one way to define it in the range we want. It seems to me that solving the above equation gives us:
$$x = e^{\log(y)/y} = y^{1/y}$$
or
$$ 1/x = (1/y) ^{1/y}.$$
This can be written in terms of the Lambert W function to give:
$$y = e^{-W(-\log(x))}.$$
I don't suppose we could get more explicit than that.
But your situation is much easier since you already know $y$. So your question is the same as "is $y= \zeta(2)-1$ a solution to $y^{1/y} = 1/2$." (No.)