In a finite field product of non-square elements is a square

If the characteristic of $F$ is $2$, then $x\mapsto x^2$ is an automorphism of $F$ and therefore every element is a square. So we can assume the characteristic is an odd prime.

Consider the multiplicative group $F^*$ of nonzero elements; the map $x\mapsto x^2$ is a group endomorphism of $F^*$ with kernel $\{1,-1\}$. Therefore the image $H$ of this map is a subgroup satisfying $|H|=|F^*|/2$; this amounts to saying that $H$, which is the set of all squares in $F^*$, has index $2$. Therefore $F^*/H$ is a two-element group and the statement follows.

Hint: I assume, $a,b,\neq 0$ otherwise it is obvious. $F^*$ is cyclic. Suppose it is generated by $x$, you can write $a=x^n, b=x^m$, $n,m$ odd. Then $n+m$ is even.

Zero is a square, hence we may assume that both $a$ and $b$ belong to $\mathbb{F}^*$.
$\mathbb{F}^*$ is a cyclic group with order $q-1$: if the characteristic is $2$, every element of $\mathbb{F}$ is a square and there is nothing to prove. Otherwise, we may assume that $\mathbb{F}^*$ is generated by some element $g$, and in such a case the quadratic residues in $\mathbb{F}^*$ are the elements of the form $g^{\text{even}}$ and the non-quadratic residues are the elements of the form $g^{\text{odd}}$. Since $\text{odd}+\text{odd}=\text{even}$, the product of two non-quadratic residues is a quadratic residue, i.e. the Legendre symbol is multiplicative.