Here's a question: if $x = \sqrt[3]{3}$, then what is the value of $y = x^{x^{x^{\dots}}}$?

As you say, if $y$ is a solution to this, then $y \ln x = \ln y$, so that
$$
\frac{\ln y}{y} = \ln(x) = \frac{\ln(3)}{3}
$$
Now, how can we "solve this" for $y$? As it ends up, there are two solutions for $y$. The answer you are getting is the second root, $y \approx 2.47805$.

We can, in fact, conclude that the equation $x^{x^{x^{\dots}}} = 3$ has *no solution*.

So, the salient question is how do we "choose" one value of $y$ over the other?

First of all, we should decide when this sequence converges at all. For a fixed $a$, define the function $f_a(x) = a^x$. Consider the following recursive definition of a sequence:

$$
x_0 = 1\\
x_n = a^{x_{n-1}}
$$
If any one value for $y = a^{a^{a^{\dots}}}$ makes sense, it's the limit $\lim_{n \to \infty} x_n$. This has now become an analysis of fixed-point iteration.

It's relatively easy to show that we can guarantee that this sequence converges as long as near $x = 1$ (our starting point), $|f'(x)| < 1$. Because $f'(x) = \ln(a) a^x$, we find that $|f'(1)| < 1$ exactly when $a < e$. So, this sequence will necessarily converge for $a \in (e^{-1},e)$.

As it turns out, however, this is not the only situation in which the sequence converges. In fact, it was shown by Euler (see wiki page and comments below) that the sequence will converge for $e^{-e} < a < e^{1/e}$. Correspondingly, the limit of the sequence will necessarily lie in $(1/e,e)$. Because $3$ lies beyond these bounds, it cannot be the limit of such a sequence.

See also this wikipedia page.