Question: My teacher asked me this question. But I think this is wrong. Anyone found this, before me? I do not know. Anyway, Is my solution correct?

${x^{x^{x^{x^x}}}}^{...}=2$

$x^2=2$

$x=\sqrt 2$

${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=2$

Now,

Let, ${x^{x^{x^{x^x}}}}^{...}=4$

$x^4=4$

$x=\sqrt2$

${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4$

Contradiction!

Answer: 
It's false: this number is defined as the limit of the sequence:
$$a\_0=\sqrt2, \quad a\_{n+1}={\sqrt{\rule{0pt}{1.8ex}2}}^{\,a\_n}$$
It is easy to show by induction this sequence is increasing and bounded from above by $2$. Thus, by the *monotone convergence theorem*, it tends to a limit $\ell\le 2$, which satisfies the equation
$$\ell=\sqrt{\rule{0pt}{1.8ex}2}^{\,\ell}\iff \log\ell=\frac{\ell\log 2}2\iff\frac{\log \ell}\ell=\frac{\log 2}2$$
It is known this equation has two solutions: $\ell=2$ or $\ell=4$. As $\ell\le 2$, the limit is actually the former.
The problem is that the expression $x^{x^{\dots}}$ is ill-defined. What you want to do is take the limit of

$$x,x^x,x^{x^x},x^{x^{x^x}},\dots$$

For $x=\sqrt2$, this limit is indeed $2$. The reason you get $4$ is because the limit of the sequence

$$4,x^4,x^{x^4},x^{x^{x^4}},\dots$$

is $4$ when $x=\sqrt2$, and appears to look the same as the other sequence, however, they are not the same.