Tensor product of fields over integers

Here is a self-contained argument. First, as Jeremy Rickard observes, $K \otimes K \cong K \otimes_k K$, where $k$ is the prime subfield of $K$ (so $\mathbb{Q}$ if $K$ has characteristic zero and $\mathbb{F}_p$ if $K$ has characteristic $p$). If $K \otimes_k K$ is a field, then as Denis Nardin observes, the multiplication map

$$K \otimes_k K \xrightarrow{m} K$$

must be injective, and since it's surjective it must be an isomorphism. But $K$ has dimension $1$ as a $K$-vector space, while $K \otimes_k K$ has dimension $\dim_k K$. It follows that $\dim_k K = 1$, hence that $K = k$.

I already wrote this in the comments but I think this might be worth of an answer. I think we can classify all fields $K$ such that $K\otimes K$ is a field.

Claim If $K$ is a field such that $K\otimes_\mathbb{Z}K$ is a field then the multiplication map $K\otimes_\mathbb{Z} K\to K$ is an isomorphism

In fact the multiplication map is always a surjection because $x\otimes 1$ gets sent to $x$. Moreover all maps of fields are injections, so the map is a bijection. But a bijective map of rings is an isomorphism.

Now the rings $R$ such that the multiplication map $R\otimes_\mathbb{Z}R\to R$ is an isomorphim have been classified by Bousfield and Kan in their paper The core of a ring. For our purposes we need lemmas 3.6 and 3.7 in said paper.

Now there are two possibilities for our field $K$: either it is of characteristic $p$ or it is of characteristic 0. If it is of characteristic 0 it cannot have any torsion subgroup. Hence from lemma 3.7 it needs to be a localization of $\mathbb{Z}$. Since it is a field it must be $\mathbb{Q}$.

If the characteristic of $K$ is $p$ it coincides with its own torsion subgroup. Hence from lemma 3.6 it is of the form $\mathbb{Z}/n$ for some $n$. Since $K$ is a field it must be $n=p$.