Example 6 (1987 Canadian Mathematics Competition) 1987 can be written as a three-digit number $\overline{x y z}$ in base $b$, if $x+y+z=1+9+8+7$, try to determine all possible $x, y, z$ and $b$.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
It is known that $x b^{2}+y b+z=1987, x+y+z=25$, thus, $x\left(b^{2}-1\right)+y(b-1)=1962$, which means $(b-z)[(b+1) x+y]=1962=2 \times 3^{2} \times 109$.
From $b>10$ we know $b-1>9$; from $1962 \geqslant b^{2}-1$ we know $b \leqslant \sqrt{1963}<45$. Therefore, $9<b-1<44$.
Since $1962=2 \times 3^{2} \times 109$, it has 12 positive divisors, which are $1,2,3,6,9,18,109,218,327,654,981,1962$.
Therefore, $b-1=18, b=19$.
From $1987=5 \times 19^{2}+9 \times 19+11$ we know $x=5, y=9, z=11$.