Just to augment Kevin's series of comments: I think that the conductor of the induction of some character $\chi$ over a quadratic field to $\mathbb Q$ would normally equal $D N(C)^2$, where $D$ is the discriminant of the quadratic field, $C$ is the condutor of the character (an ideal in the quadratic field) and $N$ is the norm from the quadratic field to $\mathbb Q$. E.g. in Kevin's $23$ example, one inducing a character of conductor 1 from $\mathbb Q(\sqrt{-23})$, so the conductor is $23$. [Added in response to an edit in the question: This form has nebentypus equal to the Legendre symbol mod 23.]
In the Maass case one should be able to do something similar, by e.g. choosing a prime $p \equiv 1 mod 4$ such that $\mathbb Q(\sqrt{p})$ has non-trivial class group, and then inducing a non-trivial character of conductor 1. [Added in response to an edit in the question: Such examples will have nebentypus equal to the Legendre symbol mod p, I think.]