Why isn't $\mathbb{CP}^2$ a covering space for any other manifold? This is one of those perhaps rare occasions when someone takes the advice of the FAQ and asks a question to which they already know the answer.  This puzzle took me a while, but I found it both simple and satisfying.  It's also great because the proof doesn't use anything fancy at all but it's still a very nice little result.
 A: Euler characteristic is multiplicative, so (since $\chi(P^2)=3$ is a prime number) if $P^2\to X$ is a cover, $\chi(X)=1$ and $\pi_1(X)=\mathbb Z/3\mathbb Z$ (in particular, X is orientable). But in this case $H_1(X)$ is torsion, so (using Poincare duality) $\chi(X)=1+\dim H_2(X)+1>1$.
A: Here's another argument that has the disadvantage of being less elementary, but the advantage of working on all $\mathbb{C}P^{2k}$ simultaneously.  (This also answers Pete's question in the comments).
We're going to apply the Lefshetz fixed point theorem which states the following:  Suppose $f:M\rightarrow M$ with $M$ "nice enough" (certainly, this applies to compact manifolds - I think it applies to all compact CW complexes).  Then $f$ induces a (linear) map $f_*:H_*(M)/Torsion\rightarrow H_*(M)/Torsion$.  Let $Tr(f)\in\mathbb{Z}$ denote the trace of this map.  If $Tr(f)\neq 0$, then $f$ has a fixed point.
Now, we'll show that every diffeomorphism $f:\mathbb{C}P^{2k}\rightarrow \mathbb{C}P^{2k}$ has trace $\neq 0$, so that every diffeomorphism has a fixed point.  Believing this for a second, note that every element of $\pi_1(X)$ for a hypothetical space $X$ covered by $\mathbb{C}P^{2k}$ acts by diffeomorphisms, and thus has a fixed point.  But it is easy to show that the only element of the deck group which fixes any point must be the identity.  It follows that $\pi_1(X)$ is trivial, so $X=\mathbb{C}P^{2k}$.
So, why does every diffeomorphism of $\mathbb{C}P^{2k}$ have a fixed point?  Well, every diffeomorphism (or even homotopy equivalence!) must act as multiplication by $\pm 1$ on each of the $2k+1$ $\mathbb{Z}$s in the cohomology ring of $\mathbb{C}P^{2k}$ and the trace  of the induced map is the sum of all the $\pm 1$s.  But since there is an odd number of $\pm 1$s, they can't sum to 0 (by, say, checking the parity), so by the Lefshetz fixed point theorem, every diffeomorphism (or even homotopy equivalence) must have a fixed point.
What about $\mathbb{C}P^{2k+1}$?  Now we must investigate using the ring structure of $\mathbb{C}P^{2k+1}$.  Since there is a single multiplicative generator, once we know what happens on $H^2(\mathbb{C}P^{2k+1})$ we know what happens everywhere.  It's easy too see that every orientation preserving homotopy equivalence must have a fixed point: if $f$ is orientation preserving, it's the identity on $H^{4k+2}(\mathbb{C}P^{2k+1})$, which implies it must have been the identity on $H^2(\mathbb{C}P^2)$ so it's the identity on all cohomology groups.  Thus, the trace of such an $f$ is $2k+1\neq 0$, and so, by the Lefshetz theorem, this map has a fixed point.
As an immediate corollary, if $\mathbb{C}P^{2k+1}$ covers anything, it can only double cover it.  For the product of any two nontrivial elements in the deck group must be trivial: any nontrivial map must be orientation reversing and the composition of two orientation reversing maps is orientation preserving, hence has a fixed point, hence is the identity.  That is, any two nontrivial elements product to $e$.  It's easy to show that this implies the Deck group is $\mathbb{Z}/2\mathbb{Z}$ (or trivial).
In fact $\mathbb{C}P^{2k+1}$ does double cover something (though, to my knowledge, it doesn't have a more common name, except in the case of $\mathbb{C}P^1 = S^2$ double covering $\mathbb{R}P^2$).  In homogeneous coordinates, the involution maps $[z_0:z_1:...:z_{2k+1}:z_{2k+2}]$ to $[-z_1:z_0:...:-z_{2k+2}:z_{2k+1}]$.  This involution acts freely, and the quotient of $\mathbb{C}P^{2k+1}$ by the involution is a space which $\mathbb{C}P^{2k+1}$ double covers.
I do not know if $\mathbb{C}P^{2k+1}$ covers anything else.
Incidentally, just to preempt a bit, the space $\mathbb{H}P^{n}$ doesn't cover anything unless $n=1$.  The proof is much more complicated in general (though the case where $n$ is even follows precisely as it did in the $\mathbb{C}P^{2k}$ case).
In general, one needs to compute Pontrjagin classes and note that they are preserved by diffeomorphisms.
We have $p_1(\mathbb{H}P^n) = 2(n-1)x$ where $x$ is a particular choice of generator for $H^4(\mathbb{H}P^n)$.  Since any diffeomorphism must preserve $p_1$, it follows that so long as $n\neq 1$, we must have $x\rightarrow x$ on $H^4$.  Then, the Lefshetz theorem once again guarantees a fixed point.
