Perhaps the following two examples would be of interest to you; my apologies if they are too simple.

Given $f \in \mathbb{F}_{q}[X_1, \ldots, X_n]$ let us define a sequence $N_s = |(H_{f} (\mathbb{F}_{q^s})|$, where $H_f(K) :=${$(x_1, \ldots, x_n) \in \mathbb{A}^{n}_{K} | f(x_1, \ldots, x_n) = 0$}.

The zeta function is then defined for a hyperplane $H_f$ and field $\mathbb{F}_q$ by

$$Z(T) = \exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big)$$

Before giving a few examples of the rationality of $Z(T)$, we recall the Maclaurin series

$$-\log(1 - T) = \sum_{s=1}^{\infty}T^s / s$$

Example 1. $f(x_1, \ldots, x_n) \equiv 0$. Then $N_s =|{\mathbb{A}}_{\mathbb{F}_{q^s}}^{n}| = q^{ns}$, so that we find $Z(T)$ becomes

$$\exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big) = \exp\big(\sum_{s=1}^{\infty} (q^n T)^s /s\big) = \exp(-\log(1-q^n T)) = 1/(1 - q^n T)$$

Example 2. Let $f = x_1 x_4 - x_2 x_3 - 1$. We now consider two cases:

Case 1. $x_3 = 0$. Then $x_1 x_4 - x_2 x_3 = 1$ becomes $x_1 x_4 = 1$. Since $x_2$ is out of the equation, it can be any element of $\mathbb{F}_{q^s}$. Thus, there are $q^s$ choices for $x_2$. Meanwhile, $x_1$ can be any nonzero element of $\mathbb{F}_{q^s}$, and in each case this will determine $x_4$. Hence there are $q^s(q^s - 1) = q^{2s} - q^s$ points in $H_f$ when $x_3 = 0$.

Case 2. $x_3 \neq 0$. Then $x_1$ and $x_4$ can be any elements of $F_{q^s}$, and $x_3$ can be any nonzero element of $F_{q^s}$. But this completely determines $x_2$, so that there are $q^s q^s (q^s - 1) = q^{3s} - q^{2s}$ points in $H_f$ when $x_3 \neq 0$.

Therefore, $N_s = q^{3s} - q^{2s} + q^{2s} - q^{s} = q^{3s} - q^{s}$, whence the zeta-function $Z(T)$ is

$$\frac{\exp(\sum_{s=1}^{\infty}q^{3s}T^s /s)}{\exp(\sum_{s=1}^{\infty}q^s T^s /s)} = \frac{1 - qT}{1 - q^3 T}$$

Note: The case for an affine variety $H_{f_1, \ldots, f_m}$ follows from the affine hypersurface $H_f$ case by a simple application of the Inclusion/Exclusion Principle. Bearing this in mind, it shouldn't be too hard to construct examples similar to those above over a variety for which the rationality can be witnessed directly. I imagine this would be somewhat tedious, though.

1

Perhaps the following two examples would be of interest to you; my apologies if they are too simple.

Given $f \in \mathbb{F}_{q}[X_1, \ldots, X_n]$ let us define a sequence $N_s = |(H_{f} (\mathbb{F}_{q^s})|$, where $H_f(K) :=${$(x_1, \ldots, x_n) \in \mathbb{A}^{n}_{K} | f(x_1, \ldots, x_n) = 0$}.

The zeta function is then defined for a hyperplane $H_f$ and field $\mathbb{F}_q$ by

$$Z(T) = \exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big)$$

Before giving a few examples of the rationality of $Z(T)$, we recall the Maclaurin series

$$-\log(1 - T) = \sum_{s=1}^{\infty}T^s / s$$

Example 1. $f(x_1, \ldots, x_n) \equiv 0$. Then $N_s =|{\mathbb{A}}_{\mathbb{F}_{q^s}}^{n}| = q^{ns}$, so that we find $Z(T)$ becomes

$$\exp\big(\sum_{s=1}^{\infty} N_s T^s /s\big) = \exp\big(\sum_{s=1}^{\infty} (q^n T)^s /s\big) = \exp(-\log(1-q^n T)) = 1/(1 - q^n T)$$

Example 2. Let $f = x_1 x_4 - x_2 x_3 - 1$. We now consider two cases:

Case 1. $x_3 = 0$. Then $x_1 x_4 - x_2 x_3 = 1$ becomes $x_1 x_4 = 1$. Since $x_2$ is out of the equation, it can be any element of $\mathbb{F}_{q^s}$. Thus, there are $q^s$ choices for $x_2$. Meanwhile, $x_1$ can be any nonzero element of $\mathbb{F}_{q^s}$, and in each case this will determine $x_4$. Hence there are $q^s(q^s - 1) = q^{2s} - q^s$ points in $H_f$ when $x_3 = 0$.

Case 2. $x_3 \neq 0$. Then $x_1$ and $x_4$ can be any elements of $F_{q^s}$, and $x_3$ can be any nonzero element of $F_{q^s}$. But this completely determines $x_2$, so that there are $q^s q^s (q^s - 1) = q^{3s} - q^{2s}$ points in $H_f$ when $x_3 \neq 0$.

Therefore, $N_s = q^{3s} - q^{2s} + q^{2s} - q^{s} = q^{3s} - q^{s}$, whence the zeta-function $Z(T)$ is

$$\frac{\exp(\sum_{s=1}^{\infty}q^{3s}T^s /s)}{\exp(\sum_{s=1}^{\infty}q^s T^s /s)} = \frac{1 - qT}{1 - q^3 T}$$