is inert in $K$ then only one of the $\theta_i$ can be ramified at $\ell$ (see [@Kraus Lemme 1]), hence the statement. Then we can restrict to the case when both characters are ramified at a prime dividing $\ell$.

If $\ell = \id{l}_1 \id{l_2}$ splits, we can assume that $\theta_i$ has conductor $\id{l}_i$ since by [@Kraus Lemme 1] they cannot both be ramified at the same prime. Then on one hand the restriction of $\theta_i$ to the inertia group at $\id{l}_i$ is the cyclotomic character, so $\theta_i(-1)=-1$, and on the other hand it is a character of $({{\mathcal{O}}}_K/\id{l}_i)^\times$ which is trivial in ${{\mathcal{O}}}_K^\times$ (as it factors through the Artin map), so $\theta_i(-1)=1$. This is impossible since $\ell\not=2$.

Lastly, suppose that $\ell$ ramifies in $K$, so $\ell \equiv 3 \pmod
  4$ and $K=\Q(\sqrt{-\ell})$. In particular, the restriction of the cyclotomic character to $\operatorname{Gal}(\overline{K}/K)$ has (odd) order $(\ell-1)/2$. Let $\id{l}$ denote the prime of $K$ dividing $\ell$ and $I_{\id{l}}$ the associated inertia subgroup. Then the $\theta_i$ are both characters of level $1$ at $\id{l}$ (since characters of level $2$ have irreducible image), $\theta_1 \theta_2
  = \chi_{\ell}$ and $E$ has good supersingular reduction at $\ell$. In the notation of [@Serre2], let $a_\ell$ denote 