Is $\mathbf{C}$ the algebraic closure of any field other than $\mathbf{R}$?

Let $S$ be a transcendence basis for $\mathbf{C}$ over $\mathbf{Q}$. Then $\mathbf{C}$ is the algebraic closure of $\mathbf{Q}(S)$. You can choose $S$ to contain any transcendental number, so take one in $\mathbf{C}\setminus\mathbf{R}$, such as $\pi\sqrt{-1}$, and then $\mathbf{Q}(S)\neq\mathbf{R}$. Also $\mathbf{Q}(S)$ can't contain $\sqrt{-1}$, so $\mathbf{Q}(S)\neq\mathbf{C}$.

Consider a transcendental extension of $\mathbb C$, $\Bbb C(t)$. Since $\sqrt t\notin\Bbb C(t)$ it is not algebraically closed and therefore the fields are different. Clearly $\Bbb C(t)$ is not isomorphic to $\Bbb R$ as well.

The algebraic closure of $\Bbb C(t)$ is of cardinality $2^{\aleph_0}$ and is therefore isomorphic to the complex numbers. This follows from the fact that the theory of algebraically closed fields in a fixed characteristics is categorical for uncountable cardinalities, that is to say once we chose the characteristics of the field there is one model up to isomorphism. So every algebraically closed field of characteristics zero whose cardinality is $2^{\aleph_0}$ must be isomorphic to the complex numbers.

Other examples of non-$\Bbb R$ fields whose algebraic closure is isomorphic to $\Bbb C$ include the $p$-adic numbers, and any other field of characteristic zero and cardinality continuum.