How do you prove that for?
x^2-1)(y^2-4)+8xy=x^2y^2-(4x^2+y^2-4xy)+4xy+4 and this is (xy+2)^2-(2x-y)^2, so we can factor this as (xy+2+2x-y)(xy+2-2x+y) \tag{1} So this should be a prime or the opposite of a prime. Then one of the factors of (1) must be \pm 1. Now the first factor of (1) is (x-1)(y+2)+4, so if this is \pm 1, then (x-1)(y+2)=-3 or -5. If (x-1)(y+2)=-3, then x-1 is \pm 1 or \pm 3 and y+2 follows from this. That gives (x,y)=(2,-5),(0,1),(4,-3),(-2,-1). Then the second factor of (1) is respectively -17,3,-21,7. So here we get the primes 3,7,17. If (x-1)(y+2)=-5, then x-1 is \pm 1 or \pm 5 and y+2 follows from this. That gives (x,y)=(2,-7),(0,3),(6,-3),(-4,-1). Then the second factor of (1) is respectively -23,5,-31,13. Combining with 1. we obtain the primes 3,5,7,13,17,23,31. One could hope that the second factor of (1) produces some additional primes, but note that if we replace x,y in the first factor by their opposite, then we get the second factor (and vice versa), so that actually will lead to the same primes.