Example 5 Find all positive integers $w, x, y, z$ that satisfy $w!=x!+y!+z!$.
Solution: Without loss of generality, let $w>x \geqslant y \geqslant z$.
If $y>z$, then dividing both sides of the equation by $z!$ yields
$$
\begin{array}{l}
w(u-1) \cdots(z+1) \\
=x \cdots(z+1)+y \cdots(z+1)+1 .
\end{array}
$$

Here, $z+1>1$ can divide the left side of the above equation, but cannot divide the right side, leading to a contradiction.
If $x>y=z$, then we have
$$
w(w-1) \cdots(z+1)=x \cdots(z+1)+2 \text {, }
$$

which implies $(z+1) \mid 2$, thus $z+1=2$. The equation can then be simplified to
$$
w(w-1) \cdots 3=x \cdots 3+1,
$$

which is also impossible.
Therefore, it must be that $x=y=z$, in which case the original equation becomes $w!=3 \cdot x!$.
$$
\therefore x=y=z=2, w=3 .
$$