Example 24 (15th Canadian Mathematical Olympiad) Find all positive integer solutions to the equation $w!=x!+y!+z!$.

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Let's assume $x \geqslant y \geqslant z > 0$. According to the problem, we have $w > x$, thus $w \geqslant x + 1$. Therefore, $(x + 1)! \leqslant w! = x! + y! + z! \leqslant 3x!$,
which implies $x + 1 \leqslant 3$, so $x \leqslant 2$. Hence, $x = 1$ or $x = 2$.
(1) If $x = 1$, then $y = z = 1$. In this case, $w! = 1! + 1! + 1! = 3$, and there is no solution for $w$.
(2) If $x = 2, y = 2, z = 2$, then $w! = 2! + 2! + 2! = 6 = 3!$, so $w = 3$.
(3) If $x = 2, y = 2, z = 1$, then $w! = 2! + 2! + 1! = 5$, and there is no solution for $w$.
(4) If $x = 2, y = 1$, then $z = 1$. In this case, $w! = 2! + 1! + 1! = 4$, and there is no solution for $w$.
Therefore, the original equation has only one unique solution: $x = 2, y = 2, z = 2, w = 3$.