Are there only two smooth manifolds with field structure: real numbers and complex numbers? Is it true that in the category of connected smooth manifolds equipped with a compatible field structure (all six operations are smooth) there are only two objects (up to isomorphism) - $\mathbb{R}$ and $\mathbb{C}$?
 A: Finite-dimensional manifolds are locally compact, and the only non-discrete locally compact topological fields are the reals and the complex numbers.  So the answer is yes.
A: Here is a series of standard arguments.
Let $(\mathbb{F},+,\star)$ be such a field. Then $(\mathbb{F},+)$ is a finite-dimensional (path-)connected abelian Lie group, hence $(\mathbb{F},+) \cong \mathbb{R}^n \times (\mathbb{S}^1)^m$ as Lie groups. Since $\mathbb{F}$ is path-connected, there is in particular a path $\gamma: [0,1] \to \mathbb{F}$ with $\gamma(0) = 0_{\mathbb{F}}$ and $\gamma(1) = 1_{\mathbb{F}}$. Now consider the homotopy $H: \mathbb{F} \times [0,1] \to \mathbb{F}$, $(x,t) \mapsto \gamma(t) \star x$. This gives a contraction of $\mathbb{F}$ and so we can exclude all the circle factors.
Now, fix $y_0 \in \mathbb{F}$ and consider the map $\widehat{y_0}: \mathbb{R}^n \to \mathbb{R}^n$, $x \mapsto x \star y_0$. Then $\widehat{y_0}$ is an additive map (but at the moment not necessarily linear with respect to the natural vector space structure on $\mathbb{R}^n$). It is not too difficult to see that by additivity we have $\forall q\in \mathbb{Q}: \widehat{y_0}(qx) = q \widehat{y_0}(x)$. Since $\widehat{y_0}$ is continuous (as being smooth), it now follows that it's actually $\mathbb{R}$-linear.
Thus $\mathbb{F}$ is an $\mathbb{R}$-algebra. From this point on one can finish either by the Frobenius theorem on the classification of finite-dimensional associative $\mathbb{R}$-algebras or invoke a theorem of Bott and Milnor from algebraic topology that $\mathbb{R}^n$ can be equipped with a bilinear form $\beta$ turning $(\mathbb{R}^n,\beta)$ into a division $\mathbb{R}$-algebra (not necessarily associative) only in the cases $n=1,2,4,8$.
EDIT: Another finishing topological argument is a theorem of Hopf saying that $\mathbb{R}$ and $\mathbb{C}$ are the only finite-dimensional commutative division $\mathbb{R}$-algebras. This is less of an overkill compared to invoking Frobenius or Bott–Milnor as the proof is a rather short and cute application of homology, see p.173, Thm. 2B.5 in Hatcher's "Algebraic Topology".
