On every other finite field at least one of −1, 2 and −2 is a square, because the product of two non squares is a square 
[Except on field extensions of $\mathbb{F}_2$] On every other finite field at least one of $−1$, $2$ and $−2$ is a square, because the product of two non squares is a square.

I don't see why this is so. Please explain. This statement is from the Wikipedia article on factorisation of polynomials over finite fields. 

Looking at the Wikipedia article, I think you meant to say "On every finite field of characteristic other than $2$" rather than "On every other finite field apart from $\mathbf{F}_2$". Also, the Wikipedia article doesn't mean to imply that the statement isn't true in finite fields of characteristic $2$—just that the characteristic $2$ statement is not needed for the argument in that particular passage.  In fact, every element in a field of characteristic $2$ is a square.  In particular, $-1=1$ and $2=-2=0$, both of which are obviously squares.
In the odd characteristic case, if you are asking how the statement that at least one of $-1$, $2$, and $-2$ is a square follows from the statement that the product of two non-squares is a square, then suppose that all three of $-1$, $2$, and $-2$ were non-squares.  Then there's an immediate contradiction, since $(-1)(2)=-2$, and therefore $-2$ should be a square.
Others have given nice proofs that the product of two non-squares is a square.  Also, once you know the fundamental fact that all nonzero elements of a finite field are powers of some primitive element $r$, it's straightforward to see this: if an element were a non-square, it would have to be an odd power of $r$.  But the product of two odd powers of $r$ is an even power of $r$.
