4-dimensional cohomology $\mathbb{CP}^2$'s

No. If $\Sigma$ is any homology 4-sphere with non-trivial fundamental group, $\mathbb{CP}^2 \# \Sigma$ is a homology $\mathbb{CP}^2$ with non-trivial fundamental group. (Here $\#$ denotes connected sum.) There are many examples of homology 4-spheres: for instance, Kervaire produced many examples in his paper Smooth homology spheres and their fundamental groups.**

Whether there is any simply connected 4-manifold with the same homology as $\mathbb{CP}^2$ but that is not diffeomorphic to it, is an important open question, which is somewhat related to the 4-dimensional smooth Poincaré conjecture.

** The main statement is about homology $n$-spheres for $n$ strictly larger than 4, but the remark at the end of the introduction mentions that the constructive part of the proof works for $n = 4$ as well.

This was originally going to be a comment to Marco Gallo's answer, but got too long. I figured since this provides an alternative solution to the problem, I'll post it as an answer.

An alternative answer to this question are the fake projective planes, these are algebraic surfaces which have the same cohomology as $\mathbb{CP}^2$, but nontrivial fundamental groups. They have universal cover the complex unit ball which topologically is just $\mathbb{R}^4$.

I offer this additional answer since they provide examples very different to Marco Galla's answer and answer a question of Anubhav Mukherjee's in the comment. Either they are irreducible or have an irreducible summand that is a cohomology $\mathbb{CP}^2$.

Suppose $X$ is one of the fake projective planes. If $X$ is irreducible, we're done. If $X$ is reducible, then it admits a connect sum decomposition $X = Y \# \Sigma$ where $Y$ is a cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ that is not a homotopy $S^4$. Again if $Y$ is irreducible and not homeomorphic to $\mathbb{CP}^2$, we're done. If $Y$ is reducible, repeat this process again until you get either an irreducible cohomology $\mathbb{CP}^2$. This process must terminate by compactness.

This gives a connect sum decomposition which we'll write as $X = Y \# \Sigma$ (slightly abusing notation here) where $Y$ is an irreducible cohomology $\mathbb{CP}^2$ and $\Sigma$ is a homology $S^4$ which is the connect sum of all of the homology $S^4$'s we found. If $Y$ is not $\mathbb{CP}^2$, we're done.

Now we see that $Y$ can not be $\mathbb{CP}^2$. Suppose $Y = \mathbb{CP}^2$ and consider the universal cover $f: \tilde{\Sigma} \rightarrow \Sigma$. Take a point $p \in \Sigma$ and it's inverse image $f^{-1}(p)$. We can now construct a cover $f^*: Y \# |f^{-1}(p)|\mathbb{CP}^2 \rightarrow \Sigma \# \mathbb{CP}^2 = X$ by connect summing a $\mathbb{CP}^2$ at each point in $f^{-1}(p)$. This covering space is simply connected since it's summands are simply connected and so it is the universal cover. This universal covering space has 2nd homology coming from the $\mathbb{CP}^2$ summands. This contradicts $X$ having universal cover $\mathbb{R}^4$. Thus $Y$ is not $\mathbb{CP}^2$ and is an irreducible cohomology $\mathbb{CP}^2$.