Since the coefficients are real, nonreal roots must come in conjugate pairs. Hence, if there is only one real root that is a root, its multiplicity must be either 2 or 4. If the multiplicity of $ r $ is 4, then $ r $ must be 2 or $ -2, $ so the quartic must be either $ (x - 2)^4 $ or $ (x + 2)^4 $. We can check that neither of these fit the given form. Hence, the quartic must be of the form $ (x - r)^2 (x^2 + bx + c), $ where $ b^2 - 4c < 0 $. Expanding, we get \[x^4 + (b - 2r) x^3 + (r^2 - 2br + c) x^2 + (br^2 - 2cr) x + cr^2 = x^4 + kx^3 + x^2 + 4kx + 16.\]Matching coefficients, we get \begin{align*} b - 2r &= k, \\ r^2 - 2br + c &= 1, \\ br^2 - 2cr &= 4k, \\ cr^2 &= 16.\end{align*}Then $ c = \frac{16}{r^2} $. Comparing $ b - 2r = k $ and $ br^2 - 2cr = 4k, $ we get \[4b - 8r = br^2 - \frac{32}{r}.\]Then $ 4br - 8r^2 = br^3 - 32, $ so $ br^3 + 8r^2 - 4br - 32 = 0 $. This equation factors as \[(r - 2)(r + 2)(br + 8) = 0.\]If $ br + 8 = 0, $ then $ b = -\frac{8}{r}, $ and \[b^2 - 4c = \frac{64}{r^2} - 4 \cdot \frac{16}{r^2} = 0,\]so this case is impossible. Therefore, either $ r = 2 $ or $ r = -2 $. If $ r = 2, $ then $ c = 4, $ $ b = \frac{7}{4}, $ and $ k = -\frac{9}{4}, $ and the quartic becomes \[x^4 - \frac{9}{4} x^3 + x^2 - 9x + 16 = (x - 2)^2 \left( x^2 + \frac{7}{4} x + 4 \right).\]If $ r = 2, $ then $ c = 4, $ $ b = -\frac{7}{4}, $ and $ k = \frac{9}{4}, $ and the quartic becomes \[x^4 + \frac{9}{4} x^3 + x^2 + 9x + 16 = (x + 2)^2 \left( x^2 - \frac{7}{4} x + 4 \right).\]Hence, the possible values of $ k $ are $ \boxed{\frac{9}{4}, -\frac{9}{4}} $.