Finite algebra over a ring is a field 
Let $F\supseteq R$ be a field and a subring. $F$ is a finite $R$-algebra. Can we conclude that $R$ is also a field?

I cannot find any counter-examples. Finite $\mathbb{Z}$-algebra (even finitely generated) that is a field must be finite. $\mathbb{Q}(\sqrt{2})$ is finite over $\mathbb{Q}$. $\mathbb{Q}(\pi)$ is not finite over $\mathbb{Q}[\pi]$.
An associated problem:

For an integral domain $R$, can $\text{Frac}(R)$ be a finite $R$-algebra and not $R$?

 A: Both of these problems can be attacked with the same tool, namely the theory of integral extensions. Since $F$ is a finite $R$-algebra, the extension $F/R$ is integral. Indeed, for any $x \in F$, multiplication by $x$ is an $R$-module endomorphism $F \to F$ which must satisfy some monic polynomial relation over $R$ by Cayley-Hamilton; since $F$ is a faithful $R$-module, this means that $x$ satisfies the same monic polynomial relation over $R$. (See, e.g., Proposition 5.1 of Atiyah-Macdonald). 
We may then prove your result. Let $R, F$ be integral domains with $R$ a subring of $F$ and $F$ integral over $R$. If $F$ is a field, then $R$ is a field. Indeed, let $x$ be a nonzero element of $R$. Then $x^{-1} \in F$, and therefore is integral over $R$, and so satisfies some monic polynomial relation over $R$ of (say) degree $d$. Multiplying this equation by $x^{d-1}$ shows that $x^{-1}$ is an $R$-linear combination of powers of $x$, hence belongs to $R$. (Conversely, if $R$ is a field, then so too is $F$; see Proposition 5.7 of Atiyah-Macdonald for details.) 
Applying this with $F = \mathrm{Frac}(R)$, we find that $R$ must be a field, and so $R = F$.  
A: The answer is yes, even if we instead of assuming that $F$ is a finite $R$-algebra we assume that it is an integral $R$-algebra. See U. Görtz, T. Wedhorn: Algebraic Geometry I, Lemma 1.9:

Let $A$ and $B$ be integral domains and let $A\rightarrow B$ be an injective integral
  ring homomorphism. Then $A$ is a field if and only if $B$ is a field.

