Let $f$ be a Maass form on the upper half-plane with nebentypus $\chi$. It is known that
$$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$L(f\times f,s)$ has a pole at $s=1$};$$
$$\text{$f$ is dihedral}\qquad\Longleftrightarrow\qquad\text{$L(\mathrm{sym}^2f,s)$ has a pole at $s=1$}.$$
On the other hand,
$$L(f\times f,s)=L(\mathrm{sym}^2f,s)L(\chi,s),$$
and the left-hand side can only have a simple pole at $s=1$, so we conclude that
$$\text{$f$ is self-dual}\qquad\Longleftrightarrow\qquad\text{$f$ is dihedral}\quad\text{or}\quad \text{$\chi$ is trivial}.$$