Find all $n \in \mathbb{N}^{*}$ such that $n^{3}+n-2$ is a power of 2.
Let $n$ be a potential solution. Then $n^{3}+n-2=(n-1)\left(n^{2}+n+2\right)$, so $n-1$ and $n^{2}+n+2$ are powers of 2. Since $n-1<n^{2}+n+2$, it follows that $n-1 \mid n^{2}+n+2$. We deduce that $n-1 \mid n^{2}+n+2-(n-1)(n+2)=4$. The positive divisors of 4 are $1,2,4$, so the only possible values for $n$ are $2,3,5$. By testing each possibility, only 2 and 5 lead to solutions.