Consider multiplying the given equation by $2.$ That gives us \[4 \cdot 2^2 + 3 \cdot 2^3 + 4 \cdot 2^4 + \dots + n \cdot 2^{n+1} = 64.\] Subtracting the original equation from this equation, we have \[n \cdot 2^{n+1} - 2^n = 64 - 32 = 32.\] Factoring the left side gives us $2^n(n+1).$ Because $n$ must be positive, that means \[2^n(n+1) = 32 = 2^5.\] Checking powers of 2, we quickly find that $2^4\cdot 5$ works, meaning $n=\boxed{20}.$
