Suppose that the integers x,y satisfy
x2−2xy+126y2−2009=0.
Looking at this as a quadratic function of x,
Δ=4y2−4×(126y2−2009)=500(42−y2)+36
should be a square number.
If y2>42, then Δ<0. So y2<42, when y2∈{0,12,22,32}, Δ∈{8036,7536,6036,3536} is not a square number.
For y2=42, Δ=500(42−y2)+36=62. According to y=4, one can get x=1 or 7; according to y=−4, one can get x=−1 or −7.
All the integer solutions are (x,y)=(1,4),(7,4),(−1,−4),(−7,−4).