Why do I get one extra wrong solution when solving $2-x=-\sqrt{x}$?

This is because the equation $\;\sqrt x=x-2$ is not equivalent to $x=(x-2)^2$, but to $$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$ Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$, hence in the present case, $x-2 \ge 0$, i.e. $x$ must be at least $2$.

Squaring can change the set of solutions

Consider $x=3$ and $x^2=9$

Such an interesting question! let's do a 'backwards' reasoning.

It's true that $x=1$ satisfy $x^2-5x+4=0$, and then clearly $x=\left(x-2\right)^2$, since $1 = (1-2)^2=(-1)^2$.

The problem appears when you take square roots, since it is not still true that $\sqrt{(-1)^2}=-1$, in fact, $\sqrt{x^2}= |x|$ so in this case $\sqrt{(-1)^2}=|-1|=1$, which is not equal to $x-2$ when $x=1$.