I want to find all triples of natural numbers $k, m$ and $n$ satisfy the equation $$2 \times k! = m! - 2 \times n!$$
My thoughts on this equation: the expression in the right side must be positive $m! - 2n! > 0$. Then $m > 2n$. What is the next step?
We want to find $k,m,n$ such that $2k!+2n!=m!$. Clearly $k\le m-1$ and $n\le m-1$. But for $m>4$, we also have $4(m-1)!<m!$, so $2k!+2n!<m!$.
Therefore all solutions must have $m \le 4$, and you can find these by hand easily enough.