Find all positive integer solutions $(x, y, u, v)$ for the equation
$$
\left|2^{x} \cdot 3^{y}-2^{u} \cdot 5^{v}\right|=2
$$
First, find the positive integer solutions of the equation
$$
2^{x} \cdot 3^{y}+2=2^{u} \cdot 5^{v}
$$
(1)
When $x \geqslant 2$, it is easy to see that $u=1$, and equation (1) becomes
$$
2^{x-1} \cdot 3^{y}+1=5^{v}.
$$

Taking both sides modulo 4, we know $x \geqslant 3$, and taking modulo 3, we know $v$ is even. Let $v=2 t$, then
$$
5^{2 t}-1=\left(5^{t}+1\right)\left(5^{t}-1\right)=4 \times \frac{5^{t}+1}{2} \times \frac{5^{t}-1}{2},
$$

so $\frac{5^{t}+1}{2}=3^{y}, \frac{5^{t}-1}{2}=2^{x-3}$. When $y=1$,
$$
t=1, \quad v=2, \quad x=4.
$$

When $y>1$,
$$
5^{2 t}-1=\left(5^{2}-1\right)\left(5^{2(t-1)}+5^{2(t-2)}+\cdots+1\right),
$$

and
$$
0 \equiv 5^{2(t-1)}+5^{2(t-2)}+\cdots+1 \equiv t \quad(\bmod 3),
$$

so $5^{2 t}-1$ is divisible by $5^{3}-1$, but $5^{3}-1=124=4 \times 31$ has a prime factor 31, which contradicts equation (2). When $x=1$, equation (1) becomes
$$
3^{y}+1=2^{u-1} \cdot 5^{v}.
$$

Taking both sides modulo 5, we know $y=4 k+2$, and taking modulo 4, we know $u=2$. So equation (3) becomes
$$
3^{y}+1=2 \cdot 5^{v}.
$$

Taking both sides modulo 4, we know $y$ is even. Let $y=2 h$, and taking modulo 5, we know $h$ is odd, then
$$
3^{y}+1=9^{h}+1=10\left(9^{h-1}-9^{h-2}+\cdots+1\right)=10 \times 5^{v-1}.
$$

If $v>1$,
$$
0 \equiv 9^{h-1}-9^{h-2}+\cdots+1 \equiv 1+1+\cdots+1 \equiv h \quad(\bmod 5).
$$

Thus, $9^{h}+1$ is divisible by $9^{5}+1$. But $9^{5}+1$ has a prime factor 1181. Therefore, $v=1, y=2$.
Therefore, the positive integer solutions of equation (1) are $(x, y, u, v)=(1,2,2,1),(4,1,1,2)$.
Next, find the positive integer solutions of the equation
$$
2^{x} \cdot 3^{y}-2=2^{u} \cdot 5^{v}
$$

When $x \geqslant 2$, $u=1$, and equation (5) becomes
$$
2^{x-1} \cdot 3^{y}-1=5^{v}.
$$

Taking both sides modulo 4, we know $x-1=1, x=2$, then equation (5) becomes
$$
2 \cdot 3^{y}=5^{u}+1.
$$

Taking both sides modulo 3, we know $v$ is odd. And
$$
6 \cdot 3^{y-1}=(5+1)\left(5^{v-1}-5^{v-2}+\cdots+1\right),
$$

When $v>1$, $y>1$, then
$$
0 \equiv 3^{y-1}=5^{v-1}-5^{v-2}+\cdots+1 \equiv v \quad(\bmod 3).
$$

Thus, $5^{v}+1$ is divisible by $5^{3}+1$, but $5^{3}+1=126=2 \times 3^{2} \times 7$ has a prime factor 7, which contradicts equation (6). Therefore, $v=1, y=1$.

When $x=1$, equation (5) becomes
$$
3^{y}-1=2^{u-1} \cdot 5^{v}.
$$

Taking both sides modulo 5, we know $y=4 k, 3^{4 k}-1$ is divisible by $3^{4}-1$, so $u \geqslant 5$.
When $v=1$, $3^{y}-1=\left(3^{2 k}+1\right)\left(3^{2 k}-1\right)=4 \times \frac{3^{2 k}+1}{2} \times \frac{3^{2 k}-1}{2}$. $\frac{3^{2 k}+1}{2}$ and $\frac{3^{2 k}-1}{2}$ are coprime, and the latter is even, so $3^{y}+1 \equiv 4,2(\bmod 8)$. So, $u \leqslant 4$.
When $u=4$, taking modulo 4, we know $y$ is odd, $3^{y} \equiv \pm 3(\bmod 5)$, at this time there is no solution.
When $u=3$, equation (4) becomes $3^{y}+1=2 \cdot 5^{v}$, which is equation (4), so $v=1, y=2$.
If $x=1$, then equation (5) becomes
$$
3^{y}+2=5^{v}.
$$

Taking both sides modulo 4, we get $y$ is odd. When $y>1$, taking modulo 9, we get $5^{v} \equiv 2(\bmod 9)$, so $v=6 k+5$. Taking both sides of equation (8) modulo 7, at this time $3^{y} \equiv 3,-1,-2(\bmod 7)$, and $5^{6 k+5} \equiv 2(\bmod 7), \frac{3^{2 k}+1}{2}=5, \frac{3^{2 k}-1}{2}=2^{u-3}$, $k=1, y=4, u=5$.
When $v>1$,
$$
3^{4 k}-1=80\left(81^{k-1}+81^{k-2}+\cdots+1\right),
$$

and
$$
0 \equiv 81^{k-1}+81^{k-2}+\cdots+1 \equiv 1+1+\cdots+1 \equiv k \quad(\bmod 5),
$$

so $3^{4 k}-1$ is divisible by $3^{5}-1$, but $3^{5}-1=242=2 \times 11^{2}$ has a prime factor 11, which contradicts equation (7).
Therefore, the solutions of equation (5) are $(x, y, u, v)=(1,4,5,1),(2,1,1,1)$.
In summary, the solutions of the original equation are $(x, y, u, v)=(1,2,2,1),(4,1,1,2),(1,4,5,1),(2,1,1,1)$.