The Law of Cosines states:
$c² = a² + b² - 2ab \cos(\gamma)$. For $\gamma = 60º$ or $\gamma = 120º$, this becomes:
$c² = a² + b² ± ab$
I used ChatGPT to find such numbers for $0 < a < b < 61$ and obtained the following:
$(m,n,k): \gamma = 60º: m² + n² − m⋅n = k²$
(3,8,7), (5,8,7), (5,21,19), (6,16,14), (7,15,13), (7,40,37), (8,15,13), (9,24,21), (10,16,14), (10,42,38), (11,35,31), (12,32,28), (13,48,43), (14,30,26), (15,24,21), (15,40,35), (16,21,19), (16,30,26), (16,55,49), (18,48,42), (20,32,28), (21,45,39), (21,56,49), (24,35,31), (24,45,39), (25,40,35), (28,60,52), (30,48,42), (32,42,38), (32,60,52), (33,40,37), (35,48,43), (35,56,49), (39,55,49).
$(m,n,k): \gamma = 120º: m² + n² + m⋅n = k²$
(3,5,7), (5,16,19), (6,10,14), (7,8,13), (7,33,37), (9,15,21), (9,56,61), (10,32,38), (11,24,31), (12,20,28), (13,35,43), (14,16,26), (15,25,35), (15,48,57), (16,39,49), (18,30,42), (21,24,39), (21,35,49), (22,48,62), (24,40,56), (27,45,63), (28,32,52), (30,50,70), (32,45,67), (33,55,77), (35,40,65), (36,60,84), (40,51,79), (42,48 78), (49,56,91), (55,57,97)
I also found a connection between triples for $\gamma = 60º$ and $\gamma = 120º$ as shown in my video: https://youtu.be/cVFaZThhKHE
However, I wasn't able to find a formula to compute the triples similar to the formula from Euclid for Pythagorean Triples.
