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) $