Your definition of P-point should end with "finite-to-one" rather than "injective". As it stands, it defines selective ultrafilters, not P-points.

As far as I know, the consistency of the statement "neither P-points nor Q-points exist" is still an open problem. It is known that this statement implies $2^{\aleph_0}\geq\aleph_3$.

EDIT in response to a request for a reference for the last statement: Although the statement is explicitly somewhere in the literature, I can't find it just now, so instead I'll indicate how it follows from two results that I can find, Theorems 9.25 and 9.27 in my chapter of the Handbook of Set Theory. Neither result is due to me; references are given in the chapter, a pre-publication version of which is available at http://www.math.lsa.umich.edu/~ablass/hbk.pdf .

The results involve the cardinal characteristics cov$(\mathcal B)$ [the smallest number of meager sets to cover $\mathbb R$], $\mathfrak d$ [the smallest number of functions $\omega\to\omega$ to dominate all such functions], and $\mathfrak c$ [the cardinal of the continuum]. It is provable in ZFC that
$$
\aleph_1\leq\text{cov}(\mathcal B)\leq\mathfrak d\leq\mathfrak c.
$$
Theorem 9.25 says (among other things) that if $\mathfrak d=\mathfrak c$ then there exists a P-point. Theorem 9.27 says (among other things) that if cov$(\mathcal B)=\mathfrak d$ then there exists a Q-point. So in order to have neither a P-point nor a Q-point, you'd need
$$
\aleph_1\leq\text{cov}(\mathcal B)<\mathfrak d<\mathfrak c,
$$
which implies $\mathfrak c\geq\aleph_3$.