No more machinery. Make yourself a table of irreducible polynomials of degrees up to 5 by thinking about how to recognize polynomials over $\mathbb Z_2$ with $0$ or $1$ as a root, then proceeding by doing a sieve of Eratosthenes (crossing out polynomials that are divisible by lower-degree irreducibles).
Leading off with the following.
If $g$ is a primitive element of a finite field then multiplication by $g$ takes a non-zero square to a non-zero non-square and vice versa.
So if we can represent $0$ as a sum of two non-squares, then multiplication by $g$ shows that $0$ can also be written as a sum of two squares:
$$
0=a^2+b^2.
$$
This implies that
$-b^2=a^2$ and a fortiori that $-1=(a/b)^2$ is a square. This is known to be the case in $\Bbb{F}_q$ if and only if $q\equiv1\pmod4$.
Consequently:
There exists arbitrarily large finite fields such that the element $0$ cannot be written as a sum of two non-squares of that field. More precisely, this happens in the field $\Bbb{F}_q$ whenever $q\equiv-1\pmod4$.
A more interesting result is that any non-zero element $z$ of a finite field $\Bbb{F}_q$, $q$ and odd number $>5$, can be written as a sum of two non-squares. This can be seen as follows.
Assume first that $z$ is a square. Then $g^{-1}z$ is a non-square. By the well- known result we can write it as a sum of two squares
$$
g^{-1}z=x^2+y^2.
$$
Because $g^{-1}z$ is a non-square, we can deduce that $x$ and $y$ must both be non-zero. This means that the elements $gx^2,gy^2$ are both non-squares, and
$$
z=gx^2+gy^2
$$
is a presentation of the required type.
If $z=ga^2$ is a non-square then we need the result (see e.g. Ireland and Rosen) that the equation
$$
x^2+y^2=1\qquad(*)
$$
has $q-\eta(-1)$ solutions (here $\eta$ is the unique multiplicative character of order two, so equal to the Legendre symbol in the case of a prime field). The equation $(*)$ is equivalent to
$$
a^2x^2+a^2y^2=a^2,
$$
so the equation $x^2+y^2=a^2$, too, has $q-\eta(-1)\ge q-1$ solutions. At most $4$ of those solutions have either $x=0$ or $y=0$. So if $q>5$, then we are guaranteed the existence of elements $x\neq0\neq y$ such that
$g^{-1}z=a^2=x^2+y^2$. Again, it follows that
$$
z=gx^2+gy^2
$$
is a presentation of $z$ as a sum of two non-square.
The OP noted themself that in the fields of $3$ or $5$ elements there are too few non-squares. For example in $\Bbb{F}_5$ the only non-squares are $2$ and $3$, and we cannot write either of those as sums of two non-squares.
For the case of the prime fields the elegant solution by Mikhail Ivanov is surely better than this argument.
Best Answer
Proposition 1 in Vitaly Bergelson, Andrew Best, and Alex Iosevich, Sums of Powers in Large Finite Fields: A Mix of Methods, The American Mathematical Monthly, 128:8, pp. 701-718. This has the advantage of being open access.