Is every polynomial with integral coefficients a Poincaré polynomial of a manifold?

Any connected, compact, orientable 2-dimensional manifold is a genus $g$ surface. Its Poincaré polynomial is $1+2gz+z^2$. So quadratic polynomials where the coefficient of $z$ is odd cannot be the Poincaré polynomial of any compact, orienteable 2-manifold.

Good question, which was already asked by Poincaré himself.

Here is the answer for connected (smooth, closed, oriented) 4-manifolds: Every symmetric polynomial of the form $$ q(z)= 1+ a z+ b z^2 + a z^3 + z^4 $$ is realized as the Poincaré polynomial of some 4-manifold, provided that $a, b\in {\mathbb N}_0$. Indeed, take $M_{a,b}$ equal to the connected sum of $a$ copies of $S^3\times S^1$ and $b$ copies of $CP^2$. For the computation of homology under connected sum, see this question. The end result is that $$ b_i(M\# N)=b_i(M)+ b_i(N) $$ unless $i=0$ or $i=d$, where $M, N$ are both $d$-dimensional closed oriented manifolds. (The answer given there implicitly assumes that the manifolds are closed, i.e. compact and with empty boundary.)

More generally, you can also prescribe the constant (= the highest) coefficient of the Poincaré polynomial as long as they are $=n\in {\mathbb N}$, by taking the disjoint union of $M_{a,b}$ with $n-1$ copies of $S^4$.

See also here for further obstructions in higher dimensions.