In $n$ (complex) variables, the degree $d$ polynomials are determined by their $\binom{n+d-1}{n-1}$ coefficients. For example, in degree $3$ in $3$ variables we have
$$f([X,Y,Z]) = a_1X^3 + a_2Y^3 + a_3Z^3 + a_4X^2Y + a_5X^2Z + a_6Y^2X $$ $$+ a_7Y^2Z + a_8Z^2Y + a_9Z^2X + a_{10}XYZ$$
and we want to consider how many parameters decide the complex structure of the zero level-set $f^{-1}(0) \subset \mathbb{CP}^2$. Ostensibly it's $10$ because there are $10$ coefficients.
But at least one "degree of freedom" can be removed: scaling all $10$ parameters by the same factor has no effect on the zero set. I am reading that this should define an elliptic curve, so it should be decided by just one complex parameter $\tau$. Supposedly this can be done by a change of variables in $\mathbb{CP}^n$? Is there a general way to count the number of independent "complex structure parameters"?
Note, this answer is incomplete. I can't remember all the details off the top of my head. I plan to update the answer when I have access to the notes in my office.
The space of degree $d$ homogeneous polynomials is $n + 1$ variables is $V\setminus\{0\}$ where $V = H^0(\mathbb{CP}^n, \mathcal{O}(d))$. As you've mentioned, rescaling the polynomial doesn't change the hypersurface, so we can consider $\mathbb{P}(V)$. Some of these polynomials will give rise to singular hypersurfaces. The set of all such polynomials is called the singular locus which we will denote by $S$. It is a Zariski closed set.
There is a right group action of $\operatorname{Aut}(\mathbb{CP}^n) = PGL(n+1, \mathbb{C})$ on $\mathbb{P}(V)$ given by $(f, A) \mapsto f\circ A$ which preserves the singular locus, i.e. if $f \in S$, then $f\circ A \in S$ for all $A \in PGL(n+1, \mathbb{C})$. So there is a right group action of $PGL(n+1, \mathbb{C})$ on $\mathbb{P}(V)\setminus S$. Let $f, g \in \mathbb{P}(V)\setminus S$ and denote the corresponding hypersurfaces by $X_f$ and $X_g$ respectively. If $f$ and $g$ are in the same orbit of this action, then there is $A \in PGL(n+1, \mathbb{C}))$ such that $g = f\circ A$. It follows that the automorphism $A : \mathbb{CP}^n \to \mathbb{CP}^n$ restricts to a biholomorphism $X_f \to X_g$. The converse is also true, i.e. if $A$ restricts to a biholomorphism $X_f \to X_g$, then $g = f\circ A$. Therefore, the orbits of $\mathbb{P}(V)\setminus S$ under the action of $PGL(n+1, \mathbb{C})$ are precisely the smooth degree $d$ hypersurfaces of $\mathbb{CP}^n$ which are biholomorphic via an automorphism of $\mathbb{CP}^n$.
A priori, two smooth degree hypersurfaces could be biholomorphic by a map which is not the restriction of an automorphism of $\mathbb{CP}^n$. I don't remember exactly what happens in this case. I think it either doesn't happen, or it does but it doesn't affect the dimension count. I will try to look this up tomorrow.
In any case, if I remember correctly, the action of $PGL(n+1, \mathbb{C})$ is free so $$\dim\, (\mathbb{P}(V)\setminus S)/PGL(n+1, \mathbb{C})$$ is an upper bound on the dimension of the moduli space of smooth degree $d$ hypersurfaces of $\mathbb{CP}^n$. As the action of $PGL(n+1, \mathbb{C})$ is free,
$$\dim\, (\mathbb{P}(V)\setminus S)/PGL(n+1, \mathbb{C}) = \dim\, (\mathbb{P}(V)\setminus S) - \dim\, PGL(n+1, \mathbb{C}).$$
As $S$ is a Zariski closed set, $\dim (\mathbb{P}(V)\setminus S) = \dim \mathbb{P}(V) = \dim V - 1 = \binom{n+d}{n} - 1$. As for the second term, $\dim PGL(n+1, \mathbb{C}) = \dim GL(n+1, \mathbb{C}) - 1 = (n+1)^2 - 1$. Therefore, the upper bound for the dimension of the moduli space is
$$\binom{n+d}{n} - (n+1)^2.$$
The moduli space of quartic K3's corresponds to $n = 3$ and $d = 4$ which gives
$$\binom{3 + 4}{3} - (3 + 1)^2 = \binom{7}{3} - 4^2 = 35 - 16 = 19.$$
The moduli space of quartic K3's is $19$-dimensional, so I suspect all of my suspicions above are correct, but I will try to confirm this tomorrow.
In the case you asked about, $n = 2$ (it is $2$ instead of $3$ because $\mathbb{CP}^n$ has $n+1$ projective coordinates) and $d = 3$, in which case the expression becomes
$$\binom{2+3}{2} - (2+1)^2 = \binom{5}{2} - 3^2 = 10 - 9 = 1.$$
So the moduli space of elliptic curves (you have to do some work to show that every elliptic curve is a degree three hypersurface of $\mathbb{CP}^2$) is one-dimensional.
A more elementary way to see this fact is to note that every elliptic curve is of the form $\mathbb{C}/\Lambda$ where $\Lambda$ is a lattice. The lattice can be chosen such that $\Lambda = \mathbb{Z}\oplus\mathbb{Z}\tau$ where $\tau \in \mathbb{H} = \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0\}$. Two such lattices $\mathbb{Z}\oplus\mathbb{Z}\tau$ and $\mathbb{Z}\oplus\mathbb{Z}\tau'$ define isomorphic tori if and only if $\tau' = \frac{a\tau + b}{c\tau + d}$ where
$$\begin{pmatrix} a & b\\ c & d \end{pmatrix} \in SL(2, \mathbb{Z}).$$
Therefore the moduli space of elliptic curves is $\mathbb{H}/SL(2, \mathbb{Z})$, in particular it is one-dimensional. In fact (again, if I remember correctly) it is isomorphic to $\mathbb{C}$ via the $j$-invariant.