If the roots are $r$ and $s$, use $(r-s)^2=(r+s)^2-4rs$ together with the given difference of $8$.

The sum of the two $y$-coordinates can be written as $m(r+s)+2b$. Also express $m$ and $b$ in terms of $r+s$ before using the given sum of $44$.

Algebra is the fastest method here. After letting $S$ be the sum of the intersection $x$-coordinates, the conditions reduce to $S^2-8S-20=0$.

Graph $y=x^2-8x-20$. Click the two $x$-intercepts to find the possible values of $S$.

For each possible value of $S$, calculate $m=S-4$. Keep only the result that satisfies $m>0$. As a verification, use $b=17-\frac{S^2}{4}$ and graph both $y=mx+b$ and $y=x^2-4x+1$; their intersection $x$-coordinates should differ by $8$.

Let the $x$-coordinates of the intersection points be $r$ and $s$. Equating the line and parabola gives $x^2-(m+4)x+(1-b)=0$, so $r+s=m+4$ and $rs=1-b$.

Let $S=r+s$. Since the difference between the roots is $8$, $(r-s)^2=64$. Also, $(r-s)^2=(r+s)^2-4rs$, so $64=S^2-4(1-b)$. Therefore, $b=17-\frac{S^2}{4}$.

At the intersection points, the $y$-values come from the line, so their sum is $m(r+s)+2b=mS+2b$. Because $m=S-4$, substitute this and the expression for $b$:

This simplifies to $S^2-8S-20=0$, or $(S-10)(S+2)=0$.

If $S=10$, then $m=S-4=6$. If $S=-2$, then $m=-6$, which does not satisfy $m>0$. Therefore, the required value is $6$.