Answer: $(4; 23)$, $(5; 25)$, $(6; 35)$.
It is easy to see that the numbers $1! + 505 = 506$, $2! + 505 = 507$, $3! + 505 = 511$ are not perfect squares. Further, $4! + 505 = 529 = 23^2$, $5! + 505 = 625 = 25^2$, $6! + 505 = 1225 = 35^2$. So we have three pairs $(4; 23)$, $(5; 25)$, $(6; 35)$ satisfying the problem conditions.

Show that there are no other solutions of the initial equation. Indeed, the numbers $7! + 505 = 5545$ and $9! + 505 = 363385$ are not perfect squares since they are divisible by $5$ but they are not divisible by $25$. The number $8! + 505 = 40825 = 25 \cdot 1633$ is not a perfect square since $1633$ is not a perfect square. At last, if $n \ge 10$, then the decimal representation of $n!$ ends by at least two zeros, so the decimal representation of the number $n! + 505 = m^2$ ends with the digits $05$. Therefore, this number is divisible by $5$ but is not divisible by $25$, so this number is not a perfect square.