Find the value of $k$ so that
\[3 + \frac{3 + k}{4^2} + \frac{3 + 2k}{4^4} + \frac{3 + 4k}{4^6} + \dotsb = 8.\]
Let
\[S = 3 + \frac{3 + k}{4^2} + \frac{3 + 2k}{4^4} + \frac{3 + 4k}{4^6} + \dotsb.\]Then
\[kS = k \cdot 3 + \frac{3k + k^2}{4^2} + \frac{3 \cdot 2k + (2k)^2}{4^4} + \frac{3 \cdot 4k + (4k)^2}{4^6} + \dotsb.\]Subtracting these equations, we get
\[(1 - k)S = 3 + \frac{3 - k^2}{4^2} + \frac{3 - k^2}{4^4} + \frac{3 - k^2}{4^6} + \dotsb = 3 + \frac{3 - k^2}{4^2} \cdot \frac{1}{1 - \frac{1}{16}}.\]Solving $\frac{1}{1 - \frac{1}{16}} = \frac{16}{15},$ we can compute
\[(1 - k)S = 3 + \frac{3 - k^2}{4^2} \cdot \frac{16}{15} = 3 + \frac{48(3 - k^2)}{4^2 \cdot 15} = 3 + \frac{4(3 - k^2)}{5}.\]Then
\[\frac{4(3 - k^2)}{5} = 8 - 3 = 5,\]so $3 - k^2 = \frac{25}{4}.$ Hence, $k = \boxed{\pm \frac{7}{2}}.$