18sters $* \star$ Find the positive integer solutions of the equation
$$
2^{x} \cdot 3^{y}+4=2^{u} \cdot 5^{y}
$$
If $x \geqslant 3$, then $u=2$, (1) becomes $2^{x-2} \cdot 3^{y}=5^{y}-1$.
(2) $\bmod 3$ shows that $v$ is even $2 t .5^{v}-1=25^{t}-1$ is divisible by 8, so $x \geqslant 5$.

When $x>5$, $5^{v} \equiv 1(\bmod 16)$, (3) thus $v$ is divisible by 4. $5^{v}-1=5^{4 t}-1=0(\bmod 13)$. But $2^{x-2} \cdot 3^{y}$ is not divisible by 13. Therefore, (2) has no solution.
When $x=5$, (2) becomes $8 \cdot 3^{y}=5^{2 t}-1$.
(4) $5^{2 t}-1=4 \times \frac{5^{t}+1}{2} \times \frac{5^{t}-1}{2}$. $\frac{5^{t}+1}{2}$ and $\frac{5^{t}-1}{2}$ are coprime, so $\frac{5^{t}+1}{2}=3^{y}, \frac{5^{t}-1}{2}=2 . t=1, y=1, v=2$.
If $x=2$, then (1) becomes $3^{y}+1=2^{u-2} \cdot 5^{v}$,
(5) the left side is an integer, of course $u \geqslant 2$. Therefore, when $y>1$, (5) has no solution. $y=1, v=1$.

In summary, the solutions to this problem are $(x, y, u, v)=(1,1,1,1),(2,2,3,1)$, $(5,1,2,2)$