IRL_ABooklet

The only solutions are $3!+3=3^2$ and $4!+3=3^3$. We can calculate directly that $5!+3=123$ is not a power of 3. For $n\ge 6$, we have $9\mid n!$ as 3 and 6 are two of the factors defining $n!$. But also 9 is a factor of all powers of 3 beyond $3^1$, so we would have the contradiction that 9 divides both $n!$ and $n!+3$.