It is known that Ax axiomatized the pseudofinite field F as follows: F is quasifinite [https://en.wikipedia.org/wiki/Quasi-finite_field] and F is PAC [https://en.wikipedia.org/wiki/Pseudo_algebraically_closed_field].
I read Zilber's article about how AСF(algebraically closed field ) is pseudofinite in Zariski language. It is also not difficult to notice that the equation $\exists x (x^2+1=0)$ has a solution in a finite field.
I think that after all, an algebraically closed field is pseudofinite.
It is clear that any PAC is AСF. A natural question arises: is any quasi-finite field an ACF?
I found one good problem, from the book [Serre, Jean-Pierre (1979), Local Fields, Graduate Texts in Mathematics, vol. 67, translated by Greenberg, Marvin Jay, Springer-Verlag, ISBN 0-387-90424-7, MR 0554237, Zbl 0423.12016], page 192, exersize 3d:
"Deduce that every ACF $\Omega$ admits a quasi-finite subfield E having $\Omega$ as its algebraic closure."
"I think that after all, an algebraically closed field is pseudofinite."
Certainly not. Let $\varphi$ be the sentence asserting that the definable function $x\mapsto x^2$ is surjective but not injective: $(\forall y\, \exists x\, x^2 = y)\land (\exists z\exists w\, (z\neq w\land z^2 = w^2))$. The sentence $\varphi$ is true in every algebraically closed field of characteristic $\neq 2$, but it has no finite model, because in a finite structure $M$, every surjective fuction $M\to M$ is injective.
"It is clear that any PAC is AСF."
No, the implication goes the other way. Algebraically closed fields are PAC, but there are PAC fields which are not algebraically closed.
"A natural question arises: is any quasi-finite field an ACF?"
No. In fact, the negative answer is immediate from the definition of quasi-finite. A field $K$ is quasi-finite if its absolute Galois group is isomorphic to $\widehat{\mathbb{Z}}$, the profinite completion of the integers. But the absolute Galois group of an algebraically closed field is the trivial group. More concretely, a quasi-finite field has a unique extension of degree $n$ for all $n\geq 1$, but an algebraically closed field has no finite extensions of degree greater than $1$.