We can write the system as \[\begin{pmatrix} 1 & k & -1 \\ k & -1 & -1 \\ 1 & 1 & -k \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}.\]This system has a nontrivial system exactly when the determinant of the matrix is 0. This determinant is \begin{align*} \begin{pmatrix} 1 & k & -1 \\ k & -1 & -1 \\ 1 & 1 & -k \end{pmatrix} &= \begin{vmatrix} -1 & -1 \\ 1 & -k \end{vmatrix} - k \begin{vmatrix} k & -1 \\ 1 & -k \end{vmatrix} + (-1) \begin{vmatrix} k & -1 \\ 1 & 1 \end{vmatrix} \\ &= ((-1)(-k) - (-1)(1)) - k((k)(-k) - (-1)(1)) - ((k)(1) - (-1)(1)) \\ &= k^3 - k.\end{align*}The solutions to $ k^3 - k = k(k - 1)(k + 1) = 0 $ are $ \boxed{-1,0,1} $.