If number of points on a manifold is $q^n ( [n+1]_q )$ does it imply a geometric relation to $A^n (P^n)$? Consider an algebraic manifold whose number of points is $q^n ([n+1]_q)$.  Is there a geometric relation to $A^n (P^n)$?  In particular, is there an equivalence in the Grothendieck ring of varieties or could there be a birational equivalence?
If there is no such equivalence in general, might some additional reasonable requirements on a manifold force there to be such an equivalence?  

Motivation: one can see that some examples of identities on the level of enumerating $F_q$ points can be lifted to geometric relations:
Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?
Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$
 A: 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 $(u^3,−u^2)$. "

