1. Define Exquisite Polynomials:
A polynomial $P(x, y)$ is called exquisite if it satisfies $P(x, y) = P(y, -x - y)$.

2. Homogeneous Polynomials:
Let $g(d)$ be the number of linearly independent homogeneous exquisite polynomials of degree $d$ up to a constant. We need to show that:
$$\sum_{l=0}^d g(l) = \left\lceil \frac{(d+1)(d+2)}{6} \right\rceil$$
This implies that a generic exquisite polynomial can be expressed as a linear combination of these homogeneous parts.

3. Crucial Claim:
If we let $U = x^2 + xy + y^2$, $V = xy(x + y)$, and $W = x^3 - 3xy^2 - y^3$, then $P$ is exquisite if and only if it can be written as $Q(U, V, W)$, where the degree of $U$ is $\leq 2$.

4. Implication:
- $U, V, W$ are all exquisite, so $Q(U, V, W)$ is exquisite as well.
- For the other direction, we construct $Q$ from $P$ by induction on $n$.

5. Observation:
If $y \mid P$, then $V \mid P$. This follows from $P(x, y) = P(y, -x - y) = P(-x - y, x)$.

6. Induction Base Cases:
The base cases $n \leq 3$ can be verified by hand.

7. Induction Step:
Assume $n \geq 4$. Let $n = 3k + 2j$ with $k \geq 0$ and $j \in \{0, 1, 2\}$. If $P$ is exquisite and the coefficient of $x^n$ is $a$, then:
$$P - aU^jW^k$$
is exquisite and divisible by $y$, so it is a multiple of $V$. Thus, we have:
$$P = aU^jV^k + VP'$$
where $P'$ is exquisite and has degree $n - 3$. By the inductive hypothesis, there is a $Q(U, V, W) = P'(x, y)$, which implies that $P$ can be written in that form as well. This procedure ensures that the degree of $U$ in the resulting polynomial is $\leq 2$.

8. Counting:
Observe that $V^k, V^{k-1}W, \ldots, W^k$ are all linearly independent. Therefore, $U^jV^k, U^jV^{k-1}W, \ldots, U^jW^k$ are also linearly independent. Since the crucial claim tells us that their linear combinations span the space of degree $n$ exquisite polynomials, we get:
$$g(3k + 2j) = k + 1$$
and $g(1) = 0$. From this, the conclusion follows.

9. Remark:
The characterization of exquisite polynomials remains the same if $F_p$ is replaced by any field with characteristic $\neq 3$. For fields with characteristic $3$, the exquisite polynomials are generated by $x - y$ and $xy(x + y)$. In the case of $F_3$, the answer would be:
$$|S_d| = 3^{\left\lceil \frac{(d+1)(d+3)}{4} \right\rceil}$$

$\blacksquare$

The final answer is $\boxed{ p^{\lceil (d+1)(d+2)/6 \rceil} }$