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?

HINT:
$n$  must be odd $=2m+1$(say)
$\implies p=4m^3+6m^2+3m=m(4m^2+6m+3)$
Both $m,4m^2+6m+3$ can not be $>1$
