To illustrate finding all primitive integer solutions, let us take $(p,q,r) = (106,92,1)$ In this case, all $x,y,z$ will be positive, and we may simply put them in decreasing order. There are three distinct parametrizations, each component a quadratic form in (coprime) integer variables $u,v,$ just like Pythagorean Triples...
$$ 106 u^2 + 197 uv + 92 v^2 \; , \; \; 92 u^2 -13 uv + v^2 \; , \; \; u^2 + 15 uv + 106 v^2 \; , \; \; $$
$$ 124 u^2 + 179 uv + 65 v^2 \; , \; \; 65 u^2 -49 uv + 10 v^2 \; , \; \; 10 u^2 + 69 uv + 124 v^2 \; , \; \; $$
$$ 130 u^2 + 159 uv + 49 v^2 \; , \; \; 49 u^2 -61 uv + 20 v^2 \; , \; \; 20 u^2 + 101 uv + 130 v^2 \; . \; \; $$
Take a pair $(u,v)$ and produce the triples given by each recipe above. If any triple comes out nonprimitive (some gcd > 1) just discard that. The primitive version will be constructed by one of the other recipes. Oh, and put in decreasing order. Call that triple $(x,y,z).$ Below, the triple in brackets is the first quadratic form in that recipe, such as <106,197,92>.
x y z first quad. form u v
106 92 1 < 106, 197, 92 > 1 0
124 65 10 < 124, 179, 65 > 1 0
130 49 20 < 130, 159, 49 > 1 0
338 251 8 < 130, 159, 49 > 1 1
368 203 26 < 124, 179, 65 > 1 1
395 122 80 < 106, 197, 92 > 1 1
887 412 94 < 130, 159, 49 > 2 1
919 302 172 < 124, 179, 65 > 2 1
1333 1246 8 < 124, 179, 65 > 1 3
1493 1048 46 < 130, 159, 49 > 1 3
1525 1000 62 < 106, 197, 92 > 1 3
1637 790 160 < 106, 197, 92 > 3 1
1696 613 278 < 130, 159, 49 > 3 1
1718 448 421 < 124, 179, 65 > 3 1
1915 1856 10 < 130, 159, 49 > 2 3
2155 1570 56 < 124, 179, 65 > 2 3
2270 1880 29 < 124, 179, 65 > 1 4
2320 1306 155 < 130, 159, 49 > 3 2
2434 1048 299 < 106, 197, 92 > 2 3
2450 1000 331 < 124, 179, 65 > 3 2
2504 1550 125 < 130, 159, 49 > 1 4
2504 754 523 < 106, 197, 92 > 3 2