Why do $x^{x^{x^{\dots}}}=2$ and $x^{x^{x^{\dots}}}=4$ have the same positive root $\sqrt 2$?

Let's first start with defining what we mean by the expression $a = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$: the infinite power tower is the limit of the recursion

$$a_{n+1} = x^{a_n},~~~~~~~ a_0 = x$$

If $x>1$ then $x^x > x$ so $a_{n}$ is monotonically increasing and converges iff it's bounded above (this is the case only if $e^{-e} \leq x \leq e^{\frac{1}{e}}\approx 1.44$; see e.g. Wiki:Tetration). If $x=\sqrt{2}$ then it follows by induction that $a_n \leq 2$ since $a_0 = \sqrt{2} < 2$ and

$$a_{n+1} = \sqrt{2}^{a_n} \leq \sqrt{2}^2 = 2$$

Thus $a_n$ is bounded above by $2$ (and the recursion therefore converges). Taking the limit of the recursion we get that it satsify the equation $a=x^a$. As you have noted this equation has the two solutions $a=2$ and $a=4$ (which are also the only real solutions). Since the limit is well defined only one of these two solutions are valid and since $a\leq 2$ we have that the solution $a=4$ does not describe the limit of the recursion above.

Why is there an extra solution? Note that

$$\lim\limits_{n\to\infty} a_n = a \implies a = x^a$$

is a one way implication and it's not true that $a = x^a \implies \lim\limits_{n\to\infty} a_n = a$. This is similar to

$$x=1 \implies x^2 = 1 \implies x = \pm 1$$

In this simple example it's squaring that introduces an extra solution. This happens quite a lot when manipulating equations: if not all of the steps are two-way implications then we might introduce extra solutions and we must use some other means to determine which one is the right one to choose.

---

In general when the infinite tower converges the limit is given by

$$a = -\frac{W(-\log(x))}{\log(x)}$$

where $W$ is the (principal branch of the) Lambert function. The maximum value (for $x$ where it converges) is found for $x = e^{\frac{1}{e}}$ where we have $a = e < 4$. Thus $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = 4$ has no solutions at all in real numbers so the manipulations you do to get $x = \sqrt{2}$ are not justified.