Pseudo algebraically closed field

Remarks

It is equivalent to ask the condition only for curves, that is, for varieties of dimension $1$.

Example: every algebraically closed field $K$ is PAC, since then $\bar K = K$ and every point of a nonempty variety is $K$-rational. Every infinite algebraic extension of a finite field is PAC.

Non-example: $\Q$ and $\R$ are not PAC. The curve $V$ defined by $x^2 + y^2 + 1 = 0$ is absolutely irreducible, since $x^2 + y^2 + 1$ is irreducible over $\C$, and $V$ is nonempty, since it contains $(i, 0)$. But it has no point with real coordinates, because $x^2 + y^2 + 1 > 0$ for real $x, y$.

Non-example: a finite field $\F_q$ is not PAC. For instance, the curve $y^q - y = x^{q+1} - x^2 + 1$ is absolutely irreducible, because the degree $q + 1$ of the right side is prime to the characteristic of $\F_q$. It has no $\F_q$-rational point, because $y^q - y = 0$ for $y \in \F_q$, while $x^{q+1} - x^2 + 1 = 1$ for $x \in \F_q$.