For $k\ge 6$ the RHS is $0\pmod{9}$ so $n^3=2\pmod{9}$, contradiction. So $k\le 5$. Checking the cases, the only solution is $k=3,n=2$.

This is much too easy. For $k>3$ we have $4 \mid k!$, so $4 \mid n^3 - 2$, hence $2 \mid n^3$, therefore $8 \mid n^3$, leading to $4 \mid n^3 - k! = 2$, absurd. Thus the only solution is $2^3 - 2 = 3!$.

my solution = let $k\ge 4 $ than $4$ divides $k!$ . hence $n^3=2 mod 4 $ . a contradiction !. thus $k < 4 $ and now checking for $k=1,2,3$ gives us $(n,k) = ( 2,3) $