Find all positive integers $n$ such that $n-1 \mid n^{3}+4$.

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Notice that $n^{3}+4=n^{3}-1+5$. Now, the factorization of $n^{3}-1=n^{3}-1^{3}$ gives us that $n-1 \mid n^{3}-1$. Therefore, $n-1 \mid n^{3}+4-\left(n^{3}-1\right)$, that is, $n-1 \mid 5$.

We have $n-1 \geq 0$. The positive divisors of 5 are 1 and 5, which gives the solutions $n-1=1$ and $n-1=5$, or equivalently $n=2$ and $n=6$. Conversely, we verify that these values of $n$ are indeed solutions.