Let 16p+1=x^3 which means $x\equiv1\mod2$, and \[16p=x^3-1=(x-1)\cdot(x^2+x+1).\]Since $x^2+x+1\equiv1\mod2$ and $x^2+x+1>x-1$, so the only case we have is \[x-1=16,\quad x^2+x+1=p\Rightarrow p=307,\]which is prime number.