[Solution] Since $y^{(x+y)^{2}}=x^{15(x+y)}=\left(y^{60}\right)^{15}=y^{900}$. Therefore, $y= \pm 1$ or $(x+y)^{2}=900$, i.e., $x+y= \pm 30$. Substituting $y= \pm 1$ into the second equation of the original system, we have
$$( \pm 1)^{x \pm 1}=x^{15},|x|=1,$$

but $x=-1$ is not a solution, while $x=1, y= \pm 1$ are two sets of integer solutions.
When $x+y=-30$, from the first equation of the original system, we get $x=y^{-2}$, substituting into $x+y=$ -30, we have $y^{3}+30 y^{2}+1=0$ which has no integer solutions.
When $x+y=30$, substituting $x=y^{2}$, we get $y^{2}+y-30=0$, which gives $y=5$ and $y=-6$, with corresponding $x=25$ and $x=36$, providing two more sets of integer solutions. Therefore, the original equation has four sets of integer solutions:
$$\left\{\begin{array} { l } { x = 1 , } \\ { y = 1 ; } \end{array} \quad \left\{\begin{array} { l } { x = 1 , } \\ { y = - 1 ; } \end{array} \quad \left\{\begin{array} { l } { x = 2 5 , } \\ { y = 5 ; } \end{array} \quad \left\{\begin{array}{l} x=36 \\ y=-6 . \end{array}\right.\right.\right.\right.$$