Powerful primes

We have $p^{2014}+p^{2015}=p^{2014}(p+1)$ . Since $p^{2014}$ is a perfect square of $p^{1007}$ , we need $p+1=n^2$ for certain $n\in\mathbb{N}$ . This gives $p=n^2-1=(n+1)(n-1)$ ; hence for $p$ to be prime we need $n-1=1$ giving us $n=2$ and $p=\boxed{3}$ .