Solve the equation $3^{x}=2^{x} y+1, x, y \in \mathbf{Z}^{+}$.
(Romania 2005 Selection Exam)
Analysis: With few conditions given in this problem, we need to further explore them.
Solve $3^{x}-1=2^{x} y$.
If $x \geqslant 3$, then $3^{x}-1 \equiv 0(\bmod 8) \Rightarrow 3^{x} \equiv 1(\bmod 8) \Rightarrow x$ is even.
Let $x=2 m \Rightarrow\left(3^{m}+1\right)\left(3^{m}-1\right)=2^{2 m} \cdot y$.
And $3^{m}+1-\left(3^{m}-1\right)=2$ and $3^{m}+1,3^{m}-1$ are both even,
then $3^{m}+1,3^{m}-1$ must be one a multiple of 2 to an odd power, and the other a multiple of 2 to an even power, otherwise the difference would be at least 4, which is a contradiction.
So one of them has only one 2, and the other contains $2^{2 m-1}$ factors. When $m \geqslant 3$, $2^{2 m}-1 \geqslant 3^{m}+1 \Rightarrow m=$ 1 , 2 , and $2 m \geqslant 4 \Rightarrow m=2$ so when $x=4$, $80=16 y \Rightarrow y=5 ; x=3 \Rightarrow$ no solution;
$$
x=2 \Rightarrow y=2 ; x=1 \Rightarrow y=1 .
$$

Therefore, $(x, y)=(1,1),(2,2),(4,5)$ are the solutions to the equation.
!Note: Congruence often plays the role of "creating something out of nothing" in indeterminate equations.