Square Divisors

Fact :- $\boxed{\text{Degree of each prime factor in factorization of a perfect square number is even.}}$

Hence, we want to find all factors of given number in which degree of every prime factor is even.

Hence we count as follows-

$\begin{aligned} \text{total number of factors} & = & \text{choices for power of 2}\times \\ & & \text{choices for power of 3}\times \\ & & \text{choices for power of 5}\times \\ & & \text{choices for power of 7} \\ & = & 2\times 3\times 4\times 5 \\ & = & \boxed { 120 } \end{aligned}$

*Two choices for powers of two are $2^{0}\text{ or } 2^{2}$

First, we list all possible $2n$ powers of a prime in this factorization ( because if $2n$ powers, then we have a certain prime factor $p^{2n} = (p^n)^2$ suggesting a perfect square for a certain $p^n$ );

We have a total of $2 \times 3 \times 4 \times 5\ = \boxed{120}$ divisors