None. Here's why:
x and y cannot be equal since
x²+x²=2007
2x²=2007
x²=2007/2
Thus x cannot be an integer.
Thus x² and y² are not equal.
0²=0, 1²=1, 2²=4, 3²=9, 4²=16, 5²=25, 6²=36, 7²=49, 8²=64, 9²=81
Therefore squares only end with 0,1,4,5,6, or 9.
Therefore the only way the sum of two different squares could end
with 7 is for one of them to end with 1 and the other to end with 6.
By symmetry, there is no loss in generality by supposing x² ends in
1 and y² ends in 6.
There are 4 cases to consider.
x ends in 1 or 9 and y ends in 4 or 6
Case 1: x ends in 1 and y ends in 4
x = 10u+1 and y = 10v+4, for some positive integers u and v
(10u+1)² + (10v+4)² = 2007
100u²+20u+1 + 100v²+80v+16 = 2007
100u² + 100v² + 20u + 80v = 1990
The left side is divisible by 20 but the right side isn't.
Thus Case 1 is out.
Case 2: x ends in 1 and y ends in 6
x = 10u+1 and y = 10v+6, for some positive integers u and v
(10u+1)² + (10v+6)² = 2007
100u²+20u+1 + 100v²+120v+36 = 2007
100u² + 100v² + 20u + 120v = 1970
The left side is divisible by 20 but the right side isn't.
Thus Case 2 is out.
Case 3: x ends in 9 and y ends in 4
x = 10u+9 and y = 10v+4
(10u+9)² + (10v+4)² = 2007
100u²+180u+81 + 100v²+80v+16 = 2007
100u² + 100v² + 180u + 80v = 1910
The left side is divisible by 20 but the right side isn't.
Thus Case 3 is out.
Case 4: x ends in 9 and y ends in 6
x = 10u+9 and y = 10v+6
(10u+9)² + (10v+6)² = 2007
100u²+180u+81 + 100v²+120v+36 = 2007
100u² + 100v² + 180u + 120v = 1890
The left side is divisible by 20 but the right side isn't.
Thus Case 4 is out.
--------------------
Thus there are no solutions.
Edwin
