Solution:

Write $16p + 1 = n^3,$ so $16p = n^3 - 1 = (n - 1)(n^2 + n + 1).$ Since $16p + 1$ is odd, $n$ is odd, and $n^2 + n + 1$ is odd as well. Therefore all four factors of $2$ must divide $n - 1:$ write $n - 1 = 16k,$ which gives $p = k(n^2 + n + 1).$ For $p$ to be prime we need $k = 1,$ so $n = 17.$

Then $p = 17^2 + 17 + 1 = 307,$ which is indeed prime, and $16 \cdot 307 + 1 = 4913 = 17^3.$