For each prime $p$, the numbers of the interval $n \in [p^2, p^3)$ are not solutions, because the degree of $p$ in the decomposition of the LCM is 2 and so contributes by a factor of 3 to the number of divisors. Therefore this number is not a power of 2. From Bertrand's postulate for a prime $p$ there is a prime $q$ with $p < q < 2p$, respectively $p^2 < q^2 < 4p^2 < p^3 < q^3$ for $p \ge 5$; also $3^2 < 5^2 < 3^3 < 5^3$. Therefore, each interval $[p^2, p^3)$ and $[q^2, q^3)$ for consecutive primes $3 \le p < q$ intersects. We obtained that $n < 9$ and a direct check shows that only $n = 1, 2, 3$ and $8$ are solutions. $\square$