What is the Cayley projective plane?

The geometric reason that there is no octonion projective 3-space is that the Desargues theorem holds in any projective space, and the Desargues theorem implies that the coordinate algebra is associative. (Essentially due to Hilbert, Grundlagen der Geometrie, 1899.)

Incidentally, the term "Cayley plane" is a misnomer, due to misidentification of the octonions with Cayley. John Graves discovered the octonions in December 1843, and they were rediscovered by Cayley a couple of years later. The octonion projective plane was first constructed by Ruth Moufang in the 1930s.

As I recall, the Cayley projective plane is painful to build, but it is a 2-cell complex, with an 8-cell and a 16-cell. The cohomology is Z[x]/(x^3) where x has degree 8, as you would expect. Its homotopy is unapproachable, because it is just two spheres stuck together, so you would pretty much have to know the homotopy groups of the spheres to know it. The attaching map of the 16-cell is a map of Hopf invariant one, from S^15 to S^8, the last such element.

I think the real reason that the Cayley projective plane exists is because any subalgebra of the octonions that is generated by 2 elements is associative. That is just enough associativity to construct the projective plane, but not enough to construct projective 3-space. And this is why you should not expect there to be a projective plane for the sedonions (the 16-dimensional algebra that is to the octonions what the octonions are to the quaternions), because every time you do the doubling construction you lose more, and in particular it is no longer true that every subalgebra of the sedonions that is generated by 2 elements is associative.
Mark

There is a topological obstruction.

First we have to specify what do we expect from a octonion projective space, because it is difficult to prove that something does not exists if it has not been defined.

We expect that there is a notion of (projective) subspaces of codimension k. Our first assumption is that the complement of a hyperplane is the affine octonion space. Then one can use induction (applying the same argument on the hyperplane) to give a cell structure on $\mathbb{OP}^n$. This already determines its additive cohomology, which has a generator in each degree multiple of 8, up to 8n.

Our second assumption is that a subspace of codimension k is the intersection of k hyperplanes; this intersection is then automatically transverse. But cup product agrees with intersection on transverse intersections. So if h is the fundamental class of a hyperplane, the cohomology in degree 8k is generated by $h^k$.

We conclude that the cohomology ring of $\mathbb{OP}^n$ must be $\mathbb{Z}[h]/(h^n)$.

Then a construction with Steenrod operations rules out the possibility that a space with such a cohomology can exist. Namely there is no space with cohomology ring $\mathbb{Z}[x]/(x^n)$ (with m>3) unless x has degree 2 or 4. For a proof see Hatcher, Algebraic topology, corollary 4.L.10.