Example 3 Find all non-negative integer solutions to the equation $2^{x}+2009=3^{y} \times 5^{z}$.
(2009, Middle European Mathematical Olympiad)
Obviously, $y, z$ are not both 0. Otherwise, the right side $=10$, then $(-1)^{x}-1 \equiv 0(\bmod 3)$.
Therefore, $2 \mid x$.
If $z>0$, then $2^{x}-1 \equiv 0(\bmod 5)$.
And $2^{x} \equiv 2,-1,-2,1(\bmod 5)$, so $4 \mid x$.
In summary, $x$ is even.
If $x=0$, then
$$
\begin{array}{l}
3^{y} \times 5^{z}=2^{x}+2009=2010 \\
=2 \times 3 \times 5 \times 67,
\end{array}
$$

which is impossible.
If $x=2$, then
$$
\begin{array}{l}
3^{y} \times 5^{x}=2^{x}+2009=2013 \\
=3 \times 11 \times 61,
\end{array}
$$

which is impossible.
When $x \geqslant 4$, $3^{y} \times 5^{x} \equiv 1(\bmod 8)$.
Since $3^{y} \equiv 3,1(\bmod 8), 5^{x} \equiv 5,1(\bmod 8)$, only when $3^{y} \equiv 1(\bmod 8)$ and $5^{z} \equiv 1(\bmod 8)$, we have $3^{y} \times 5^{z} \equiv 1(\bmod 8)$.
Thus, $y, z$ are both even.
Let $x=2 x_{1}, y=2 y_{1}, z=2 z_{1}$. Then
$$
\left(3^{y_{1}} \times 5^{x_{1}}+2^{x_{1}}\right)\left(3^{y_{1}} \times 5^{x_{1}}-2^{x_{1}}\right)=7^{2} \times 41 \text {. }
$$

Since $3^{y_{1}} \times 5^{x_{1}}+2^{x_{1}}>3^{y_{1}} \times 5^{x_{1}}-2^{x_{1}}$, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
3^{y_{1}} \times 5^{z_{1}}+2^{x_{1}}=49, \\
3^{y_{1}} \times 5^{z_{1}}-2^{x_{1}}=41 ;
\end{array}\right. \\
\left\{\begin{array}{l}
3^{y_{1}} \times 5^{s_{1}}+2^{x_{1}}=7 \times 41, \\
3^{y_{1}} \times 5^{s_{1}}-2^{x_{1}}=7 ;
\end{array}\right. \\
\left\{\begin{array}{l}
3^{y_{1}} \times 5^{x_{1}}+2^{x_{1}}=7^{2} \times 41, \\
3^{y_{1}} \times 5^{x_{1}}-2^{x_{1}}=1 .
\end{array}\right.
\end{array}
$$

Subtracting the two equations gives $2^{x_{1}+1}=8,280,2008$.
But only $2^{x_{1}+1}=8$ has a solution $x_{1}=2$. Hence, $x=4$.
Thus, $3^{y_{1}} \times 5^{z_{1}}=45=3^{2} \times 5$.
Therefore, $y_{1}=2, z_{1}=1$. Hence, $y=4, z=2$.
In summary, $(x, y, z)=(4,4,2)$.