x^2 + (x+2)^2=y^2 where x and y anre ints.
You can investigate only-int solutions. May be a little complicated be should not be impossible.
OR (assuming the +2 side can be any side (hypo.) too)
x^2 + y^2 = (x+2)^2 gives
y^2 = 4X + 4
y = 2 * sqrt(x+1)
x= 8 gives y=6. All are integers.
x= 15 gives y=8, and x+2=17 all integers and fits the answer.
So, there should be infinitly many solution in the above format.
Please note Solution to the first expression may be hard but I suspect if there are infinitely many solutions from: x^2 + (x+2)^2=y^2 probably only 3, 4, and 5. But you got to check that out.