can anyone please help me solve this one?
the result have to be in positive integers

Step-by-step explanation:
(a²-b²)(c²-d²) = a²c² - a²d² - b²c² + b²d²
(b²-d²)(c²-a²) = b²c² - a²b² - c²d² + a²d²
the sum of both is
a²c² + b²d² - a²b² - c²d² = a²(c²-b²) + d²(b²-c²) =
= a²(c²-b²) - d²(c²-b²) = (c²-b²)(a²-d²) = 2021
after a little check we find 2021 is only divisible by 43 and 47 (or 1 and 2021).
so, it is clear that our expression resembles
43×47 = 2021
that means that e.g. c²-b² = 43, and a²-d² = 47
or the other way around, it does not matter.
what squared integer numbers give a difference of 43 ? and 47 ?
squared numbers
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256
289 324 361 400 441 484 529 576
aha ! 484 - 441 = 43
and 576 - 529 = 47
so, we get
a = sqrt(576) = 24
b = sqrt(441) = 21
c = sqrt(484) = 22
d = sqrt(529) = 23
one could also use the formula for consecutive squared numbers
(n+1)² = n² + 2n + 1
under the assumption that the 2 numbers are convective (which it turns out they are in this example) we get
(n+1)² - n² = 2n + 1 = 43
2n = 42
n = 21
n+1 = 22
and for 47
2n + 1 = 47
2n = 46
n = 23
n+1 = 24