I am searching for an example of a curve that isnt defined over a numberfield (i.e. isn't isomorphic to an algebraic curve with coefficients in a numberfield). Because of Belyis Theorem this is equivalent to that there isnt a belyi function (a meromorphic function with at most $3$ ramification values) for this curve. Do you have such an example?
Thanks for every answer!
As suggested in the comments, the standard approach is to choose a curve with transcendental moduli. The easiest case is genus one: an elliptic curve $E$ is entirely determined up to isomorphism over $\mathbb{C}$ by its $j$-invariant. Given $j_0 \in \mathbb{C} \setminus \{0, 1728\}$, the curve $$ E: y^2 + xy = x^3 - \frac{36}{j_0 - 1728} x - \frac{1}{j_0 - 1728} $$ has $j$-invariant $j_0$. (See the proof of Prop. III.1.4 (p. 47) in Silverman's The Arithmetic of Elliptic Curves.)
So take $j_0$ to be some transcendental number, say $\pi$, in the above formula and let $E$ be the resulting elliptic curve. We claim $E$ cannot be defined over any number field. For contradiction, suppose $E$ is isomorphic to some $E'$ whose defining equation has coefficients in some number field $K$. Since $E'$ has an equation over $K$, and its $j$-invariant is a rational function in the coefficients of this equation, then $j(E') \in K$. (WLOG, we may take $E': y^2 = x^3 - 27 c_4 x - 54 c_6$ with $c_4, c_6 \in K$, so $j(E') = 1728 \frac{c_4^3}{c_4^3 - c_6^2} \in K$.) But since $E$ and $E'$ are isomorphic, then $j(E') = j(E) = \pi \notin K$, contradiction.