jmc

If $n$ is a perfect cube, then all exponents in its prime factorization are divisible by $3$. If $n$ is a perfect fourth power, then all exponents in its prime factorization are divisible by $4$. The only way both of these statements can be true is for all the exponents to be divisible by $\mathrm{lcm}[3,4]=12$, so such an $n$ must be a perfect twelfth power. Since we aren't using $1^{12}=1,$ the next smallest is $2^{12}=\boxed{4096}.$