66th Czech and Slovak Mathematical Olympiad

Suppose that positive integer $a$ satisfies $p^n+1=a^3$ (clearly $a\ge 2$). We rewrite the equality as $$p^n=a^3-1=(a-1)(a^2+a+1).$$ It follows that if $a>2$, the numbers $a-1$ and $a^2+a+1$ are powers of $p$ (with positive integer exponents). If $a>2$ then $a-1=p^k$, hence $a=p^k+1$ for some positive integer $k$. Plugging this into $a^2+a+1$ gives $p^{2k}+3p^k+3$. Since $a-1=p^k<a^2+a+1$, the trinomial $a^2+a+1$ is a higher power of $p$, and thus $$p^k\mid p^{2k}+3p^k+3\Rightarrow p^k\mid 3.$$ Then $p=3$ and $k=1$, hence $a=p^k+1=4$. However, number $a^2+a+1=21$ is not a power of three therefore if $a>2$, the expression $p^n+1$ is never a cube. For $a=2$ we get $p^n=7$, hence $p=7$ is the only such prime.