The trouble is that the squaring function cannot be inverted "cleanly", because if we have $x^2 = 4$, then there are two possibilities for what $x$ could be (namely $2$ and $-2$). Thus if we have $x^2 = a^2$, then we can remove the two squares, but we must account for the possibility of negative solutions by manually saying $x = \pm a$.

Looking at your particular case: $(1)^2 = (-1)^2$ we can apply the same logic to conclude
$$ 1 = \pm(-1) = \pm 1 $$
which is true. $1$ *does* in fact equal $1$ **or** $-1$ (in this case we obviously know it's $+1$, but that doesn't mean that the plus-or-minus expression is wrong, just kind of silly).

So the short answer is, you cannot undo the squares without introducing (manually) the plus-or-minus into the expression. After you do that, you may be able to determine *which* of them it is, but that's your own job.

As for your other question, why isn't it that $x^2 = 4 \implies \pm x = \pm 2$? Well, actually, I guess it *does*, but it's unnecessary to put the $\pm$ on *both* sides. You only need to put it on one. I remember myself once wondering why does $\pm$ get placed on only one side? Don't you need to do it to *both* sides? But this is a misunderstanding of what is going on when we make the logical leap from $x^2 = 4$ to $x = \pm 2$. You might think we are taking the $\pm \sqrt{\ \ } $ of both sides, but that is not really true. $\pm \sqrt{\ \ }$ is not a *function*, so the usual rule of applying it to both sides doesn't really make sense. Really all we're doing is just remembering that there is always two solutions to a quadratic, and thus we use that principle (and no other) when we deduce that $x^2 = 4 \implies x = \pm 2$.