Solution:

We can rewrite $4096$ as $2^{12},$ so $\dfrac{1}{4096} = 2^{-12}.$ Then if we equate the given expressions, we get $$2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}.$$ Equating the exponents, we get $$m - 6 = 1 + \dfrac{-12}{m}.$$

Multiplying by $m,$ we get $$m^2 - 6m = m - 12$$ and so $$m^2 - 7m + 12 = 0$$$$(m-4)(m-3)$$$$m=4,~m=3$$

Therefore, we can see that the sum of the solutions is $7.$

Thus, C is the correct answer.