Example 5 Two boxes of candy have a total of 176 pieces. 16 pieces are taken from the second box and placed into the first box, at which point, the number of pieces of candy in the first box is 31 more than $m($ an integer $m>1)$ times the number of pieces of candy in the second box. Then, the first box originally had at least $\qquad$ pieces of candy.
Let the first box originally contain $x$ candies, and the second box originally contain $y$ candies.
According to the problem, we set up the system of equations:
$$
\left\{\begin{array}{l}
x+y=176, \\
x+16=m(y-16)+31 .
\end{array}\right.
$$

Rearranging, we get
$$
x+16=m(176-16-x)+31 \text {, }
$$

which simplifies to $x=\frac{160 m+15}{m+1}=160-\frac{145}{m+1}$.
Since $x$ and $m$ are both positive integers, we have:
$$
\begin{array}{l}
160-\frac{29 \times 5}{m+1} \\
\geqslant 160-29=131(m>1) .
\end{array}
$$

Therefore, the first box originally contained at least 131 candies.