Yes, it’s always true.

If we call our first number a^2+b^2 and our second number c^2+d^2, since we are told that each is the sum of different squares we can, without loss of generality, specify that a>b and c>d. It is possible for a=c or for b=d, but not both at the same time, since we are told the two sums of squares are different.

If we multiply out (a^2+b^2)(c^2+d^2), we get (ac)^2+(ad)^2+(bc)^2+(bd)^2, which is the sum of four squares, not what we are after. Our approach will be to call the target product (e^2+f^2) and to express e and f in terms of a,b,c,d and demonstrate that (e^2+f^2) is equivalent to (ac)^2+(ad)^2+(bc)^2+(bd)^2.

This can be achieved by letting e=ad+bc and f=ac-bd

e^2+f^2 = (ad)^2+2abcd+(bc)^2+(ac)^2-2abcd+(bd)^2

Clearly the abcd terms cancel out, leaving only the terms we were hoping for.

However, this isn’t quite enough to satisfy the question, since e^2 and f^2 need to be different, and our expressions do not guarantee that. For instance if, as in the original example, a=2, b=1, c=3, d=1, both e and f are equal to 5.

Instead let us call g=ac+bd and h=ad-bc (whereas before f was guaranteed to be positive, h now could be negative, but it doesn’t really matter, as h^2, which is what we are really interested in, will be positive regardless).

Now g^2+h^2 = (ac)^2+2abcd+(bd)^2+(ad)^2-2abcd+(bc)^2

Again the abcd terms cancel and leave just what we want. But now, since a>b and c>d, g^2 will always be bigger than h^2, meaning we always have two different squares summing to the target value. This method might yield h^2 = 0, which is technically a square number, but if we would prefer non-zero square numbers we can revert to the e and f equations, knowing that if h = 0, e cannot be equal to f. I have a nice proof of this, but I won’t go into it now.