Find all prime numbers p such that 2p+1 is a perfect cube. $$2p+1=n^3$$
$$ 2p = n^3 - 1$$
$$ 2p = (n-1)(n^2+n+1)$$
The only number that fits the criteria is 13, how can I prove this?

Your final equation has $2p$ written as a product, and you already know there are very few ways to do this for $2p$ -- namely, one of the factors have to be either $1$ or $2$. So just check each of the cases
$$ n-1 = 1 \quad\text{or}\quad n-1=2 
\quad\text{or}\quad n^2+n+1 = 1 \quad\text{or}\quad n^2+n+1 = 2 $$
