Five, (17 points) Find the positive integer solutions of the equation
$$
x^{2}+6 x y-7 y^{2}=2009
$$
Five, factoring on the left side, we get
$$
(x-y)(x+7 y)=2009 \text{. }
$$

Since $2009=7 \times 7 \times 41$, we have:
When $x-y=1,7,41,49,287,2009$, correspondingly, $x+7 y=2009,287,49,41,7,1$.
Also, since $x, y$ are positive integers, then
$$
x-y<x+7 y \text{. }
$$

Therefore, among the six relationships above, only three can hold:
(1) $\left\{\begin{array}{l}x-y=1, \\ x+7 y=2009 ;\end{array}\right.$
(2) $\left\{\begin{array}{l}x-y=7, \\ x+7 y=287\end{array}\right.$
(3) $\left\{\begin{array}{l}x-y=41, \\ x+7 y=49 .\end{array}\right.$

Solving these, we get $(x, y)=(252,251),(42,35),(42,1)$.
These are all the positive integer solutions to the equation.