jmc

If we fix $b$ then increasing $n$ increases the number of factors, so we want $n$ to equal $15$. Recall that the number of prime factors of $p_1^{e_1}{p}_2^{e_2}\cdots p_m^{e_m}$ equals $(e_1+1)(e_2+1)\cdots (e_m+1)$, where the $p_i$ are primes. Thus we want the exponents in the prime factorization of $b$ to be as large as possible. Choosing $b=12=2^2\cdot 3$ gives $e_1=2,e_2=1$. Any other number less than or equal to $15$ will either be prime or will be the product of two primes, giving smaller exponents in the prime factorization. Thus $b=12$ is the best choice, and we have $b^n=2^{30}{3}^{15}$, which has $(30+1)(15+1)=\boxed{496}$ positive factors.