Since $\Bbb C$ is an algebraic closure of $\Bbb R$, and algebraic closures are unique up to isomorphism over the base field, that makes $\Bbb C$ the unique algebraic extension up to $\Bbb R$-isomorphism. That is, if we have two algebraic closures $K_1$ and $K_2$ of a field $k$, there exists a $K_1\xrightarrow{\sim} K_2$ which restricts to $k\to k$.
It is certainly not the only extension - we have ways of constructing extensions arbitrarily, indeed there are extensions of any field of every possible cardinality, so in particular the isomorphism classes of extensions is too big to even be a set!
One learns there are two basic types of extensions: algebraic and transcendental. The first occurs when one adjoins elements that satisfy algebraic relations, the second occurs when one adjoins transcendentals. The field $\Bbb R(X)$ of rational functions in an indeterminate $X$ has the element $X$ which satisfies no algebraic relation over $\Bbb R$ (since distinct rational functions are distinct elements of $\Bbb R(X)$), so this is such an example. Any extension can be split into an algebraic extension of a purely transcendental extension.
In general, extensions are not unique. The rationals $\Bbb Q$ have infinitely many algebraic extensions that lie between it and its algebraic closure $\overline{\Bbb Q}$. Some of them are isomorphic but distinct even, for instance $\Bbb Q(\sqrt[3]{2})\cong\Bbb Q[X]/(X^3-2)\cong \Bbb Q(e^{2\pi i/3}\sqrt[3]{2})$ (the isomorphisms preserve $\Bbb Q$) but these two fields $\Bbb Q(\sqrt[3]{2})\ne\Bbb Q(e^{2\pi i/3}\sqrt[3]{2})$ are not equal since one has complex numbers and the other doesn't.