The Russell Cubic $R:=V(x + x^2 y + z^2 + t^3)\subset \mathbb{A}^4$ is not isomorphic to $\mathbb{A}^3$ although over $\mathbb{C}$ they are both diffeomorphic to $\mathbb{R}^{6}$ (see this Wikipedia page).
I ran a Mathematica program I quickly wrote to compute the number of solutions of $x + x^2 y + z^2 + t^3$ over $\mathbb{F}_p$. For the first twelve primes $p$ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37) I got $p^3$.
Based on this evidence I guessed that although $R$ and $\mathbb{A}^3$ are not isomorphic that they had the same counting polynomial.
In the comments below, Vladimir Dotsenko, provides an elementary proof of my guess:
Consider the zero set of $x+x^2y+z^2+t^3$. Note that for $x\not=0$ we have the unique value for $y$, so this gives $(p−1)p^2$ points ($p−1$ choice for $x$, $p$ choices for $z$, $p$ choices for $t$). For $x=0$, the polynomial becomes $z^2+t^3$, so there are $p$ choices for $y$ and a choice of a zero $(z,t)$ of that polynomial. However, it clearly has $p$ zeros by the usual parametrization of a singular cubic curve: for $t=0$ there is just $z=0$, and for $t\not=0$, we have $t=−(z/t)^2$ so denoting $z/t=u$, we have $(p−1)$ solutions $(u^3,−u^2)$.