What is p?

We are given

$16p+1=n^3\phantom{cccccc}n\in\mathbb{Z}_{>0}$

Moving the $1$ to the $RHS$ and using the difference of cubes identity

$16p=(n-1)(n^2+n+1)$

Isolating $p$

$p=\dfrac{(n-1)(n^2+n+1)}{16}$

Since $p$ is an integer, $16$ must divide $(n-1)(n^2+n+1)$ , in addition $n^2+n+1$ is always odd

$\Longrightarrow 16\mid n-1\Longrightarrow n=16k+1\phantom{ccccc}k\in\mathbb{Z}_{\geq 0}$

Substituting back

$p=k(n^2+n+1)$

If $k=0$ we get $p=0$ , which is not prime

If $k=1$ we get $p=307$ , which is prime

If $k\geq 2$ , neither $k$ nor $n^2+n+1$ is equal to $1$ , so we get that $p$ is composite

Hence $p=\boxed{307}$ is the only solution

This is an AIME problem from 2015. Alternatively, here's how I found it:

$\begin{aligned} a^3 & = 16p + 1 & \small \color{#3D99F6} \text{Taking both sides} \ \bmod 6 \\ a^3 & \equiv 1 \bmod{16} \\ a & \equiv 1 \bmod{16} \\ a & = 16q + 1\end{aligned}$

Conveniently enough, for $q = 1$ we have $a = 17$ and $p = \boxed{307}$ , which is prime.