Right Triangle's Proof

If $$144=(c+b)(c-b)$$ we know that $c+b$ is a divisor of $144$. So there are only finitely many pairs of $(c,b)$ we have to investigate. $(c+b,c-b)$ are from the set $$\{(1,144),(2,72),(3,48),(4,36),(6,24),\ldots,(144,1)\}$$. But we know that $c+b \gt c-b$ and that both $c+b$ and $c-b$ are even. Because $12\cdot 12=144$ we have $c-b<12$ and $c+b>12$. The only even divisors of $144$ are $\{2,4,6,8\}$. So your solutions are the only ones.

But the side of length $12$ can be the hypotenuse $c$. So you should check the equation $$12^2=a^2+b^2$$ too. Again only finitely many pairs are to check. You can assume that $a \ge b $ and therefore $a \ge \sqrt{\frac{144}{2}} = 8.\cdots$. So you have only to check $a \in \{9,10,11\}$.