Solution:

The product is a perfect square exactly when each of the primes $2,$ $3,$ and $5$ appears with even exponent. Only a roll of $5$ contributes the prime $5,$ so the number of $5$s is even: $0,$ $2,$ or $4.$ Classify the other values by the parities of their exponents of $2$ and $3:$ rolls of $1$ and $4$ contribute $(0, 0),$ a $2$ contributes $(1, 0),$ a $3$ contributes $(0, 1),$ and a $6$ contributes $(1, 1).$ The exponent of $2$ is even iff the count of $2$s plus the count of $6$s is even, and similarly for $3$s and $6$s, so a collection of non-$5$ rolls works exactly when the counts of $2$s, $3$s, and $6$s are all even or all odd.

With no $5$s, all four rolls come from $\{1, 2, 3, 4, 6\}.$ All-even cases: no $2$s, $3$s, or $6$s gives $2^4 = 16$ sequences (each roll is $1$ or $4$); exactly two of a single kind gives $3 \cdot \frac{4!}{2!\,2!} \cdot 2^2 = 72;$ two of each of two kinds gives $3 \cdot \frac{4!}{2!\,2!} = 18;$ four of one kind gives $3.$ All-odd case: one $2,$ one $3,$ one $6,$ and one roll from $\{1, 4\}$ gives $4! \cdot 2 = 48.$ Subtotal $16 + 72 + 18 + 3 + 48 = 157.$ With two $5$s, choose their positions in $\binom{4}{2} = 6$ ways; the other two rolls must lie in the same parity class, giving $2^2 + 1 + 1 + 1 = 7$ ordered pairs, for $42$ sequences. With four $5$s there is $1$ sequence.

In total $157 + 42 + 1 = 200$ of the $6^4 = 1296$ sequences work, so the probability is $\frac{200}{1296} = \frac{25}{162},$ and $m + n = 25 + 162 = 187.$