It is well-known that if the quotient field of a commutative noetherian integrally closed domain $R$ is algebraically closed, then $R$ is a field.
The proof is easy: let $r_0 \in R$ and choose $r_i \in R, \ i \geq 1$ such that $r_i^2=r_{i-1}$ (this is possible because the quotient field of $R$ is algebraically closed and $R$ is integrally closed). Now consider the chain $Rr_0 \subseteq Rr_1 \subseteq \cdots $. The rest of the proof is straightforward.
A more interesting fact is that we don't need to assume that $R$ is integrally closed, i.e. if the quotient field of a commutative noetherian domain $R$ is algebraically closed, then $R$ is a field. Is there an easy proof of this? (This is Exercise 4.2, page $90$ of the book Integral Closure of Ideals, Rings, and Modules, by Craig Huneke and Irena Swanson (2006)).
Assume $R$ is no field, i.e. there is some prime $P$ of height $1$. Then $S:= R_P$ is noetherian and one-dimensional. By Mori-Nagata-Theorem, the integral closure $\overline S$ is also noetherian (This is on pages 89/90 in your book). Hence we can apply the known case to $\overline S$ and get that $\overline S$ is a field. $S \subset \overline S$ is integral, hence $S$ is also a field. But $S$ was the localization at a height $1$ prime, hence not a field. Thus our assumption was false and $R$ is a field.