The count is $2$ and $6$ for $p=2$ and $p=3$ respectively,
and otherwise $p-1$ or $p+1$ according as $p$ is $1$ or $-1 \bmod 3$.

More generally, it is well-known that a smooth plane conic 
over ${\bf Z} / p {\bf Z}$ has $p+1$ points in the *projective* plane,
so we need only subtract the number of points at infinity, 
which here is the number of square roots of $-3 \bmod p$.
(The primes $p=2,3$ are special because $x^2+3y^2-1$ factors as
$(x+y-1)^2 \bmod 2$ and $(x+1)(x-1) \bmod 3$.)