Is the following statement true?
Let $F$ be a field and $n \in \mathbb{N}$, then there exists an extension $E$ of $F$ with $[E:F]=n$.
I'm thinking of $[\mathbb{C}:\mathbb{R}]=2$ and since I can not think of another extension of the real numbers, this would be a counterexample. But I do not know a rigorous argument, why there can not be other extensions of $\mathbb{R}$.
As illustrated in a comment, the result is false in general: Algebraically closed fields only have transcendental extensions, which have infinite order.
The question of whether numbers other than $2$ are possible is very interesting. Any extension of $\mathbb R$ is either transcendental, or of degree $2$ (and isomorphic to $\mathbb C$). In fact, if $[E:F]$ is finite and $E$ is algebraically closed, then $[E:F]=2$, and $F$ "looks a lot" like $\mathbb R$. Technically, $F$ is what we call a real closed field so, in the sense of model theory, it has the same first-order properties as the reals. In particular, any finite extension of $F$ is isomorphic to $E$, and $E$ is just $F(i)$, where $i^2=-1$.
This result is the Artin-Schreier theorem. A very nice, accessible presentation, can be found in this write-up by Keith Conrad.