I believe the answer is that you get such a deformation by varying which moment map fiber you take in the Hamiltonian reduction construction (and maybe these are even known to be all deformations.)

@mrw: I have edited this answer to also deal with the case $d$ even.

@joro: I think Vojta will apply to the first because you are looking at integral points, which corresponds to looking at a quartic surface with a log divisor of the hyperplane section at infinity instead of just a quartic surface. The quartic surface is Calabi-Yau, but the log quartic surface is log general type.

Do the solutions you produce lie in a closed (not necessarily irreducible) subvariety? I believe that Vojta's conjecture would imply that they would have to in both examples.

@skeptic: Yep, that gives you one direction. To get the other direction, you need to make sure that you don't have any small (i.e. lower dimension than expected) extraneous irreducible components, which is why you need that your locus is locally defined by the right number of equation.

It is maybe helpful to first think about the case where $X$ is $Y/G$, $Y$ a variety and $G$ a finite type algebraic group acting on $Y$. I believe this is how to define the cotangent bundle in this case: Let $Z$ be the locus in $T^*X$ of cotangent vectors at $x\in X$ which pair to zero with tangent vectors at $x$ coming from the infinitesimal action of $G$. Then $T^*X$ is just $Z/G.$ The remark you want to prove will follow because $Z$ is locally defined by $\operatorname{dim} G$ equations, which gives you a codimension bound on all components.

To define this locus as a substack, pass to a smooth cover of $X$ by affines. Then this dimension condition defines a locally closed subscheme, and you can descend this back to a substack. The codimension of the substack is the same as the codimension upstairs of this locally closed subscheme.

Now use the natural identification $\operatorname{Sym}^2(\mathbb{P}^1)\cong\mathbb{P}^2: (x,y)\mapsto (x+y,xy)$ (using affine coordinates here to make what is going on a bit clearer). We see that lines on the upper half plane are just the real points of self-conjugate lines in $\mathbb{P}^2.$ Unfortunately this uses that we have a very special collection of rational curves in $\mathbb{P}^2$ and so it's not clear how to generalize this picture beyond $SL_2(\mathbb{R})$.

An attempt: The (compactified) upper half plane is $SL_2(\mathbb{R})/K,$ for $K$ a maximal compact. This is the set of real points of $SL_2(\mathbb{R})/K_{\mathbb{C}},$ and as $K_\mathbb{C}$ is just the torus, this can be identified with $\operatorname{Sym}^2(\mathbb{P}^1)-\Delta,$ where $\Delta$ is the diagonal. Under the conjugation this variety inherits, the compactified upper half plane (i.e. the real points) corresponds to pairs of points which are each others complex conjugates.

The three $\mathbb{F}_9$-points are the points corresponding to $(0,i),(1,i)$, and $(2,i),$ where $i$ is a square root of $-1$ in $\mathbb{F}_9$.

An effective divisor of degree 2 either comes from a sum of two points of degree $1$ over your base field or from one point of degree $2$ over your base field. Similarly an effective divisor of degree $3$ on your curve would either be a sum of $3$ $\mathbb{F}_3$-points, a sum of a $\mathbb{F}_9$-point and a $\mathbb{F}_3$-point, or a single $\mathbb{F}_{27}$-point.

You also have divisors coming from $\mathbb{F}_9$-points, which should give you the remaining divisors counted by $a_2.$