Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$ 
Show that only one prime can be expressed as $n^3-1$ for some positive integer $n$

$n^3-1=(n-1)(n^2+n+1)$. If  $n^3-1$ is prime then $n-1=1$ or $n=2$ so $n^3-1=7$.
But I need a concrete proof of it, not from intuition.  

$$n^3-1=(n-1)(n^2+n+1)$$ Suppose that $n^2+n+1\le n-1\implies n^2\le -2$ which leads to contradiction.
Therefore $n^2+n+1\gt n-1$ and since $n^3-1$ is prime only possibility is $$n-1=1\implies \color{Green}{n=2}\,\,\,\,\,\text{and} \,\,\,\,\,\,\color{Red}{n^3-1=7}$$ 
