Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$?

$x^2 - 2^2 = 0$

$(x+2)(x-2)=0$

$x+2=0$ or $x-2=0$

$\sqrt4 = 2$

$x-\sqrt4=0$

$x=2$

Why can't the square root of $4$ be $-2$ instead of $2$, if $-2$ times $-2$ also equals $4$?

I think you are confused in the square root function, actually the think making you confused is that what is the value of $\sqrt4$ which is $2$ but not $\pm2$, since the square root of any number is positive.

And the case in which the value of $\sqrt4$ is $\pm2$ is a quadratic equation, and the thing which happens here is $$x^2-4=0$$ $$x^2=4$$ $$x=\pm\sqrt4$$ hence, $$x=\pm2$$ but not $$x=\sqrt4\ne\pm2$$ And in case of your question:

The first equation can be written as $$x-2=0$$ $$\implies x=2$$ $$OR$$ $\color{red}{\text{Since the first equation is linear so it cannot have two roots.}}$

Just because

$$\sqrt4=2$$ and the equation says

$$x=\sqrt4.$$