18.4.29 $\star \star$ Find the positive integer solutions to the equation
$$
3^{x}-5^{y}=2
$$
Solving $\bmod 4$ gives $(-1)^{x}-1 \equiv 2(\bmod 4)$. Therefore, $x$ is odd. Solving $\bmod 3$ gives $-(-1)^{y} \equiv -1(\bmod 3)$. Therefore, $y$ is even.

Clearly, $x > y \geqslant 1$. When $x=3$, $y=2$. When $x=5$, there is no solution. Assume $x \geqslant 7$. Multiplying both sides of (1) by $3^{x}$ and completing the square gives $\left(3^{x}-1\right)^{2}-3^{x} \cdot 5^{y}=1$. That is, $3^{x}-1, 3^{\frac{x-1}{2}} \cdot 5^{\frac{y}{2}}$ are solutions to the Pell equation
$$
u^{2}-3 v^{2}=1
$$

The solutions to (2) are known to be
$$
\begin{array}{l}
u_{n}=\frac{1}{2}\left((2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right) \\
v_{n}=\frac{1}{2 \sqrt{3}}\left((2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right) .
\end{array}
$$

Therefore, from (3), if $v_{n}=3^{\frac{x-1}{2}} 5^{\frac{y}{2}}(n \geqslant 5)$, then
$$
3^{\frac{x-1}{2}} \cdot 5^{\frac{y}{2}}-2^{n-1} \cdot n+2^{n-3} \mathrm{C}_{n}^{3} \cdot 3+2^{n-5} \mathrm{C}_{n}^{5} \cdot 3^{2}+\cdots
$$

Let $3^{\alpha} \| n$, then by (4), $\alpha>0$.
$$
\begin{aligned}
2^{n-1} \cdot n+2^{n-3} C_{n}^{3} \cdot 3 & =2^{n-4} n(8+(n-1)(n-2)) \\
& =2^{n-4} n\left(n^{2}-3 n+10\right) .
\end{aligned}
$$
$3 \mid n$, so $3 \nmid\left(n^{2}-3 n+10\right)$, thus the power of 3 in $2^{n-1} \cdot n+2^{n-3} \mathrm{C}_{n}^{3} \cdot 3$ is $\alpha$. For $k \geqslant 2$, $\mathrm{C}_{n}^{2 k+1} \cdot 3^{k}=\frac{n}{2 k+1} \cdot 3^{k} \mathrm{C}_{n-1}^{2 k}$ has a power of 3 at least $\alpha+1$ (the power of 3 in $2 k+1$ is less than $k-1$. For $k \geqslant 3$, $3^{k-1}>2 k+1$). Therefore, by (4), $\alpha=\frac{x-1}{2}$. But then
$$
3^{x}-1=u_{n}=2^{n}+\cdots>2^{3^{\frac{x-1}{2}}}>2^{2 x}
$$
(it is easy to show by induction that for $x \geqslant 7$, $\left.3^{\frac{x-1}{2}}>2 x\right)$. Contradiction! Therefore, $v_{n}=3^{\frac{x-1}{2}} 5^{\frac{y}{2}}(n \geqslant 5)$ does not hold.
It is also easy to see that $v_{0}=0, v_{1}=1\left(u_{0}=1, u_{1}=2\right)$, and $v_{n}=4 v_{n-1}-v_{n-2}$. So $v_{2}=4, v_{3}=15, v_{4}=56$, none of which can be written as $3^{\frac{x-1}{2}} 5^{\frac{y}{2}}(x \geqslant 7)$.

Therefore, the only solution to this problem is $(x, y)=(3,2)$.