$x^{2}+2\ge 2\sqrt 2 x$, $y^{2}+3\ge 2\sqrt 3 y$, $z^{2}+4\ge 4z$. Therefore if x,y,z is solution, then $x^{2}+2\le \frac{5\sqrt 3 }{6}x$, $y^{2}+3\le \frac{15}{2\sqrt 2 }y$, $z^{2}+4\le \frac{15}{\sqrt 6 }z.$ Therefore $x=1$ or $x=2$, $y=1$ or $y=2$ or $y=3$, $z=1$ or $z=2$ or $z=3$ or $z=4$. $y=2$ or $z=2$ or $z=3$ give not solution. All solutions (1,1,1),(1,1,4),(1,3,1),(1,3,4),(2,1,1),(2,1,4),(2,3,1),(2,3,4).

x^2+2>=3x (1), because x^2-3x+2=(x-1)(x-2)>=0. y^2+3>=4y (2), because y^2-4y+3=(y-1)(y-3)>=0, if y doesn't equal to 2.If y=2,then 7(x^2+2)(z^2+4)=120xz(3).So,x or z is divisible by 7. Without loosing generality, let x=7k.Then, by (3) we have: 120kz=(49k^2+2)(z^2+4),but z^2+4z>=4z, because z^2-4z+4=(z-2)^2>=0.So, 120kz>=(49k^2+2)4z,or 49k^2+2<=30k,which is impossible for positive integer k. z^2+4>=5z(4),because z^2-5z+4=(z-1)(z-4)>=0,unless z=2 or z=3.If z=2,15xy=(x^2+2)(y^2+3). If y=1,y=2,y=3 and y=4,easy to check that x isn't integer.So,(4) is correct. By multiplying (1),(2) and (4),we have:(x^2+2)(y^2+3)(z^2+4)=60xyz>=3x*4y*5z=60xyz. So, x^2+2=3x,y^2+3=4y,z^2+4=5z,which is the same as: x=1 or x=2,y=1 or y=3,z=1 or z=4.By checking, the solutions are;(1;1;1);(1;1;4);(1;3;1);(1;3;4);(2;1;1);(2;1;4);(2;3;1);(2;3;4);

$60xyz=(x^{2}+2)(y^{2}+3)(z^{2}+4)>(xyz)^{2}$ so $60xyz>(xyz)^{2}$ $60>xyz$ now, prove $min(x,y,z)<4$ now, it is easy to continue !