Given a compact Riemann surface $X$, every holomorphic function on $X$ is constant. This is obvious if you think about holomorphic functions as locally conformal mappings, that is, transformations of $X$ into a plane which locally preserve angles ("infinitesimal similarities"). Such a transformation is continuous, thus has a compact image, but the image is also open, so it must be constant: collapses $X$ into a point. In other words, the ring $\mathcal{O}(X)$ of holomorphic functions on $X$ reduces to the constants $\mathcal{O}(X) = \mathbb{C}$.
That's why, in the theory of compact Riemann surfaces, one is interested in meromorphic functions, i. e., those with at least one pole. These functions constitute a field denoted by $\mathcal{M}(X)$, and can be seen as branched coverings $X \rightarrow \mathbb{S}^2$ of the Riemann sphere/complex projective line.
One can show that there always exist non-constant meromorphic functions on $X$ (Riemann's existence theorem). This means that every compact Riemann surface is a branched covering of the sphere. This can be used to show that the field $\mathcal{M}(X)$ is a finite field extension of $\mathcal{M}(\mathbb{S}^2) =\mathbb{C}(z)$ (this last field is the field of rational functions = polynomial fractions).
Moreover, one can show that each finite extension of $\mathbb{C}(z)$ gives rise to a unique compact Riemann surface, up to isomorphisms, with the given extension as its field of meromorphic functions (Dedekind-Weber theory of algebraic function fields in one variable). This compact Riemann surface can always be realized as an algebraic curve in the complex projective space $\mathbf{P}^3(\mathbb{C})$ (without singularities, naturally). This means that it is the set of zeroes of some homogeneous polynomials with complex coefficients, in projective space.
One interesting question is to ask when a compact Riemann surface $X$ can be given by an equation with coefficients in $\overline{\mathbb{Q}}$ (the algebraic closure of rationals). It is known that this is the case if and only if there is a meromorphic function $X \rightarrow \mathbb{S}^2$ with at most three critical values (Belyi's theorem). A function of this kind is a covering of the sphere which is branched over three points (or less).
Thus, the study of compact Riemann surfaces/smooth plane algebraic curves over the algebraic numbers $\overline{\mathbb{Q}}$ is reduced to the study of coverings of the sphere branched over three points, which we can assume to be the points $0, 1, \infty$.
These branched coverings $f: X \rightarrow \mathbb{S}^2$ can be given a geometric representation. The fiber of $0$, is a finite set of points in $X$, which can be marked as black points. Points in the fiber of $1$ are usually colored white. The preimage of the interval $[0,1]$ (as a curve joining $0$ and $1$) is given by a set of curves joining black and white points, alternatively. This graph on $X$, formed by black points, white points and curves, is the dessin d'enfant associated to the branched covering $f: X \rightarrow \mathbb{S}^2$.
It is a remarkable fact that the branched covering $f: X\rightarrow \mathbb{S}^2$ determines the Riemann surface structure of $X$ (by pullback of the same structure in $\mathbb{S}^2$), and so, compact Riemann surfaces over algebraic numbers (with a specified meromorphic function branched over at most three points) are just orientable compact surfaces with certain graphs in them (the graphs associated with branched coverings).
Now, Grothendieck was interested in the Galois group of $\overline{\mathbb{Q}}$ over $\mathbb{Q}$, which acts on the coefficients of the equation of an algebraic curve over $\overline{\mathbb{Q}}$, giving another curve of the same type. Considering the dessins associated to those curves, that group then transforms a dessin into another dessin, and so, dessins can be used to obtain a geometric interpretation of the absolute Galois group.
Another example, like JSE's, that comes already equipped with a Belyi map but is not as familiar as modular curves and Fermat curves: For any relatively prime integers $m,n$ with $0<m<n$, and any subgroup $G$ of $S_n$, the curve that parametrizes trinomials $x^n + a x^m + b$ up to scaling with Galois group contained in $G$. The Belyi map is the invariant $a^n/b^{n-m}$ of the trinomial, and its degree is $d=[S_n:G]$; it is branched at $0$, $\infty$, and $(-n)^n/(m^m (n-m)^{n-m})$. One may assume $m \leq n/2$ (by symmetry with respect to $x \leftrightarrow 1/x$, $m \leftrightarrow n-m$). Some nontrivial examples with $n=5,7,8$ are given explicitly at http://www.math.harvard.edu/~elkies/trinomial.html; the subsequent paper with N.Bruin on the cases $(m,n) = (1,7)$ and $(1,8)$ with $d = 30$ is
Nils Bruin and Noam D. Elkies, Trinomials $ax^7+bx+c$ and $ax^8+bx+c$ with Galois Groups of Order 168 and $8 \cdot 168$, Lecture Notes in Computer Science 2369 (proceedings of ANTS-5, 2002; C.Fieker and D.R.Kohel, eds.), 172-188.
(These examples all have $G$ transitive, but the construction works for all subgroups $G$.)
Best Answer
If $X$ is a (smooth projective) curve over $\overline{\mathbb{Q}}$, we define
The Belyi degree is a function on $\mathcal{M}_g(\overline{\mathbb{Q}})$ which satisfies the following Northcott-type finiteness property.
Proposition. (Strong Northcott) For every integer $d$, the set of $\overline{\mathbb{Q}}$-isomorphism classes of curves $X$ over $\overline{\mathbb{Q}}$ with $\deg_B(X)\leq d$ is finite.
Proof. Like all finiteness statements, this one also boils down to some "general" finiteness statements. In this case, the statement (seemingly arithmetic in nature) is a consequence of a (topological) finiteness property of the fundamental group of $\mathbb{P}^1\setminus \{0,1,\infty\}$. Indeed, the proposition can be proven using the fact that the fundamental group of $\mathbb{P}^1\setminus \{0,1,\infty\}$ is finitely generated, and that a finitely generated group has only finitely many finite index subgroups of index at most $d$. QED
Note that this proposition can be used to enumerate all (isomorphism classes of) curves over $\overline{\mathbb{Q}}$. Simply "write" down the curves of Belyi degree at most $3$, then $4$, then $5$, etc.
The Northcott property satisfied by the Belyi degree is much stronger than that of any Weil height $h$. The Northcott property for a Weil height usually requires in addition a bound on the degree of the point.
The Strong Northcott property implies that, given a Weil height $h$ (or any function!) on $\overline{\mathbb{Q}}$, there is a function $f(\deg_B(-))$ such that
$$ h(X) \leq f(\deg_B(X)).$$
Thus, any function on $\ {\mathcal{M}_g}(\overline{\mathbb{Q}})$ is bounded by a function in the Belyi degree (simply because of the above proposition). For example, the genus of $X$ is bounded by $\deg_B(X)$. This follows from the Riemann-Hurwitz formula.
There are a few natural (arithmetic) invariants on $\ {\mathcal{M}_g}(\overline{\mathbb{Q}})$ such as the Faltings height for which one can write down explicit bounds. For example:
Theorem. If $X$ is a curve over $\overline{\mathbb{Q}}$ with Faltings height $h_F(X)$, then $$h_F(X) \leq 10^8 \deg_B(X)^6.$$
This (with many more explicit inequalities) is proven in [1]. The motivation for proving such inequalities is that they can be used to control the running time of certain algorithms computing coefficients of modular forms.
The question of actually computing the Belyi degree of a curve is an interesting one. An algorithm (which I would not recommend trying to implement) for doing so is given in [2].
[1] A. Javanpeykar. Polynomial bounds for Arakelov invariants of Belyi curves, with an appendix by Peter Bruin. Algebra and Number Theory, Vol. 8 (2014), No. 1, 89–140.
[2] A. Javanpeykar and J. Voight. The Belyi degree of a curve is computable Contemp. Math., 2019, 722, p. 43-57.