A theorem in Shaum's Outline of Linear Algebra says that if we suppose the field $K$ is infinite then any system of linear equations has either a unique solution, or no solution, or an infinite number of solutions.
So if we have $n$ unknowns then solutions are vectors in $K^n$ which can be real or complex.
Why do we want $K$ to be infinite? Wouldn't any subset of $K$ still ensure the trichotomy i.e. we would still have either a unique solution, or no solution, or an infinite number of solutions?
As pointed out in a comment, if $K$ is a finite field, then $K^n$ has $|K|^n$ elements so there cannot be an infinite number of solutions.
The trichotomy then breaks up into $n+2$ cases. The linear system $Ax=b$ may have no solution - this occurs when $b \notin \mathrm{im}A$. On the other hand, if the solution set of $Ax=b$ is non-empty then it is a linear subspace of $K^n$. For every $d=0, 1, \dots, n$ it is possible that the solution set is $d$-dimensional and thus has $|K|^d$ elements.
Note that when $K$ is infinite, the above $n+2$ cases still appear, but the solution set is infinite for $d=1, \dots, n$.