Three, (25 points) Given that the difference between the product and the sum of two positive integers is equal to twice the sum of their greatest common divisor (GCD) and least common multiple (LCM). Find all such pairs of positive integers.

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Let the required positive integers be $x, y$, and their greatest common divisor be $m\left(m \in \mathbf{N}_{+}\right)$. Then
$$
x=m a, y=m b\left(a, b \in \mathbf{N}_{+},(a, b)=1\right) \text {. }
$$

Thus, the least common multiple of $x, y$ is $m a b$.
According to the problem, we have
$$
m a \cdot m b-(m a+m b)=2(m+m a b) .
$$

Therefore, $(m-2) a b=2+a+b$.
So, $0<m-2=\frac{2+a+b}{a b}=\frac{2}{a b}+\frac{1}{a}+\frac{1}{b} \leqslant 4$.
Solving, we get $2<m \leqslant 6$.
Thus, $m=3,4,5,6$.
(1) When $m=3$, equation (1) becomes $a b=2+a+b$.

So, $(a-1)(b-1)=3=1 \times 3$.
Solving, we get $(a, b)=(2,4)$ or $(4,2)$.
Both solutions contradict the fact that $a, b$ are coprime, so they are discarded.
(2) When $m=4$, equation (1) becomes $2 a b=2+a+b$.

So, $(2 a-1)(2 b-1)=5=1 \times 5$.
Solving, we get $(a, b)=(1,3)$ or $(3,1)$.
In this case, the required two positive integers are $4, 12$.
(3) When $m=5$, equation (1) becomes $3 a b=2+a+b$.

So, $(3 a-1)(3 b-1)=7=1 \times 7$.
Solving, we get $(a, b)=\left(\frac{2}{3}, \frac{8}{3}\right)$ or $\left(\frac{8}{3}, \frac{2}{3}\right)$.
Both solutions contradict the fact that $a, b$ are positive integers, so they are discarded.
(4) When $m=6$, equation (1) becomes
$4 a b=2+a+b$.
So, $(4 a-1)(4 b-1)=9$.
When $(4 a-1)(4 b-1)=9=1 \times 9$, solving, we get $(a, b)=\left(\frac{1}{2}, \frac{5}{2}\right)$ or $\left(\frac{5}{2}, \frac{1}{2}\right)$.
Both solutions contradict the fact that $a, b$ are positive integers, so they are discarded.
When $(4 a-1)(4 b-1)=9=3 \times 3$, solving, we get $(a, b)=(1,1)$.
In this case, the required two positive integers are 6,6.
In summary, the two positive integers are 4,12 or 6,6.