Why is the solution to $x-\sqrt 4=0$ not $x=\pm 2$? If the equation is $x-\sqrt 4=0$, then $x=2$.
If the equation is $x^2-4=0$, then $x=\pm 2$.
Why is it not $x=\pm 2$ in the first equation?

Obviously the answer to 
$x - k = 0$ is $x = k$.  One answer.
And easily, but not obviously, the answer to $x^2 - k^2 = 0$ is $x =\pm k$.  Two answers (if we assume $k \ne 0$).  
So Question number 1:  Why does $x - k$ have one answer while $x^2 - m = 0$ has two (assuming $m > 0$)?
And obviously $x - \sqrt{4} = 0$ has solution $x = \sqrt{4}$ and $x^2 - 4=0$ has two solutions $x = \pm 2$.
So Question number 2: Why is $\sqrt{4} = 2$ and not $\sqrt{4} = \pm 2$.
The answer to Question number 1 is pretty clear:  For any $x \ne 0$ we know $(-x)^2 = (-1)^2 x^2 = x^2 > 0$ and we also know that if $0 < a < b$ then $a^2 < b^2$ i.e. $a^2 \ne b^2$.  And we also know that for and $m > 0$ that there must be some number $b>0$ so that $b^2 = m$.  (Actually, that last one is not obvious at all.  But we can take it as a given.  It's a matter of knowing the reals are in a continuum but... we'll leave it for now.)
Those three bits of knowledge allow as to conclude: for any $m > 0$ there are two possible numbers so that $x^2 = m$ one is .... let's call it $b$ and the other is $-b$.  So $x^2 - m = 0 \implies x^2 = m$ has two answers.
And $x - k =0 \implies x = k$ has only one.
So... we just have to answer Question number 2:  Why is $\sqrt{4} = 2$ and not $\sqrt{4} = \pm 2$.
The answer to that is that the $\sqrt{}$ of a number $m$ is NOT defined to be "the number that when squared is $m$".  If $m > 0$ then there are two such numbers and that definition to refer to a single number is ambiguous.
Instead the $\sqrt{}$ of a number $m$ is defined to be "the positive or zero number that when square is $m$".  If $m \ge 0$ there is only one such number.  If $m > 0$ then there are two such numbers that when squared are $m$ but only one of them is positive.
So if we have $x^2 - m = 0 \implies x^2 = m$ there are two answers.  The positive one is $\sqrt{m}$.  The negative one is $-\sqrt{m}$.  Or in other words $x^2 - m = 0 \implies x = \pm \sqrt{m}$.
So the statement "$x - \sqrt{4} = 0$ is not "$x$ minus one or the other of the two numbers that when squared are $4$ is equal to zero".  It is actually "$x$ minus one or the positive numbers that when squared is $4$ is equal to zero".  Or in other words $x - \sqrt{4} = 0 \iff x - 2 = 0$.
