Let $p$ be a prime number and let $q = p^2$. Let $C$ be a separated scheme of finite type over $\mathbb F_q$ of dimension $1$.
If we know that for every $\alpha \in \mathbb Z_{>0}$, "the number of $Spec(\mathbb F_{q^{\alpha}})$-points on $C$" $= C_1 q^{\alpha} + C_2p^{\alpha}$ for some $C_1, C_2 \in \mathbb Z$.
Then can we read off some geometric information of $C$ from the above point counting formula (for example, $C_1$ is related to the number of geometrically irreducible components of $C$)?
This formula implies that the zeta function of $C$ is given by the formula
$$\zeta_C( u) = e^{ \sum_{\alpha=1}^{\infty} | C(\mathbb F_{q^\alpha})| u^\alpha / \alpha } = e^{ \sum_{\alpha=1}^{\infty} (C_1 q^{\alpha} + C_2 p^\alpha) u^\alpha / \alpha } = \frac{1}{ (1 - q u)^{C_1} (1 - p u)^{C_2}} $$
Now, the zeta function of the resolution of singularities of $C$ differs from the zeta function of $C$ itself. However, this difference is only by factors of the form $1/ (1 - u^n)$. For example, if we resolve a point of degree $n$ to obtain two points of degree $n$, we get a factor of $1/(1-u^n)$. Compactifying will similarly only multiply or divide by factors of this type.
Thus the zeroes and poles of the zeta function at points satisfying $|u|<1$ will be unchanged by the resolution process.
We know the zeta function of a smooth projective geometrically irreducible curve of genus $g$ has a pole of order $1$ at $u= q^{-1}$, a pole of order $1$ at $u = 1$, and zeroes of total order $2g$ with absolute value $\sqrt{q}^{-1}$. So taking a union of curves, we see the order of the pole at $q^{-1}$ is the total number of irreducible components, and the total degree of the zeroes of absolute value $\sqrt{q}^{-1}$ is twice the sum of the genuses of the irreducible (geometric) components.
In your case, that implies $C_1$ is the number of irreducible components and $-C_2/2$ is the total genus of the components, since we have a zero of order $-C_2$ at $p^{-1} = \sqrt{q}^{-1}$ and no other zeroes on the circle of radius $\sqrt{q}^{-1}$.
