$\left(\sqrt{2}\right)^{\sqrt{2}^{{\,}^{\textstyle.^{\textstyle\,.^{\textstyle \,.}}}}}=x$
well if the infinite power tower of roots is $x$, then
$\left(\sqrt{2}\right)^x=x$
Now, I just tried out reasonable candidates, namely $2$ and $4$, which are the solution(s).
But how to do it without "brute force"?
Also, if I just type it in my calculator, $2$ seems to be the limit of this expression, so is $4$ wrong?
Let's interpret the infinite tower as the limit of the sequence $a_0 = \sqrt{2}, a_1 = \sqrt{2}^{a_0} = \sqrt{2}^{\sqrt2}, a_2 = \sqrt2^{a_1} = \sqrt2^{\sqrt2^{\sqrt2}}, \ldots$ . Indeed any limit $x$ must satisfy $x = \sqrt{2}^x$, which has the solutions $x=2$ and $x=4$.
To show that the limit must be $2$, you can prove that if $a_i \le 2$ then $a_{i+1} \le 2$. Therefore, by induction, $a_i \le 2$ for all $i$, so the limit must also be $\le 2$.