The equation of the circle is $x^2+y^2-4x-8y-5=0$
Its centre is $\left(2,4\right)$ and radius is $\sqrt{4+16+5}=5$ units
It the circle intersects the line $3x-4y=m$ at two distinct points, then the length of the perpendicular from the centre is less than the radius
i.e., $\frac{\left|6-16-m\right|}{5}<5$
$$\begin{gathered} \Rightarrow \left|10+m\right|<25\Rightarrow -25<m+10<25 \\ \Rightarrow -35<m<15 \end{gathered}$$

Hence, the total number of integral values of $m$ is $49$