8. $3+2 \sqrt{6}$.

Transform the given function to get
$$\begin{array}{l} y=\frac{m x^{2}+n x+4}{x^{2}+1} \\ \Rightarrow(m-y) x^{2}+n x+4-y=0 . \end{array}$$

Consider it as a quadratic equation in $x$, by $\Delta \geqslant 0$ we get
$$\begin{array}{l} n^{2}-4(m-y)(4-y) \geqslant 0 \\ \Rightarrow 4 y^{2}-(4 m+16) y+16 m-n^{2} \leqslant 0 . \end{array}$$

Now consider it as a quadratic inequality in $y$, given that its solution set is $y \in[1,6]$, hence
$$\left\{\begin{array} { l } { \frac { 4 m + 1 6 } { 4 } = 1 + 6 , } \\ { \frac { 1 6 m - n ^ { 2 } } { 4 } = 1 \times 6 } \end{array} \Rightarrow \left\{\begin{array}{l} m=3, \\ n=2 \sqrt{6} . \end{array}\right.\right.$$

Therefore, $m+n=3+2 \sqrt{6}$.