Let $K$ be an algebraically closed field of characteristic $0$, and $V\subset K^n$ the zero-set of $n-1$ polynomials over $K$ in $n$ variables.
Question 1. If $V$ is non-empty, can it be finite?
Question 2. What if $K$ is not algebraically closed and $V$ non-empty?
Edit. If $n=2$, then either the polynomial $p$ defining $V$ is in one variable $x$ and $V$ projects onto $K$ via the other coordinate, or the two coordinates appear in $V$, and then $V$ projects onto a cofinite subset of $K$ via either coordinates.
$V$ can be finite, since it can be empty. For instance, one of your polynomials could be $1$. If $V$ is nonempty and $K$ is algebraically closed, though, it must be infinite. This follows from basic dimension theory (essentially Krull's principal ideal theorem): each defining polynomial decreases the local dimension at any point by at most $1$, so $V$ has positive dimension at each of its points, whereas a finite set has dimension $0$ at each point. (If you're not familiar with these results, this is really a longer story than I can tell in a brief answer here, but you should be able to find this fact in some form in any introductory algebraic geometry text.)
If $K$ is not algebraically closed, then all bets are off. For instance, for $K=\mathbb{R}$, the single point $\{(0,0)\}\subset\mathbb{R}^2$ is the vanishing set of the polynomial $x^2+y^2$. In fact, every finite subset of $K^n$ is the vanishing set of a single polynomial if $K$ is not algebraically closed (see Affine variety over a field which is not algebraically closed can be written as the zero set of a single polynomial).