jmc

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 $$x^4+(b-2r)x^3+(r^2-2br+c)x^2+(b{r}^2-2cr)x+c{r}^2=x^4+k{x}^3+x^2+4kx+16.$$Matching coefficients, we get $$\begin{array}{rl}b-2r& =k,\\ r^2-2br+c& =1,\\ b{r}^2-2cr& =4k,\\ c{r}^2& =16.\end{array}$$Then $c=\frac{16}{r^2}.$ Comparing $b-2r=k$ and $b{r}^2-2cr=4k,$ we get $$4b-8r=b{r}^2-\frac{32}{r}.$$Then $4br-8r^2=b{r}^3-32,$ so $b{r}^3+8r^2-4br-32=0.$ 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 $\boxed{\frac{9}{4},-\frac{9}{4}}.$