If $k=0$ then equation $a^2+b^2=c^2$ has an infinite number of solutions in integers.
But is something special about the case $k=0$?
In other words, suppose that we choose some $k \in \mathbb Z$ and try to determine when $a^2+b^2+k=c^2$ has an infinite number of solutions.
Is it true that for every $k \in \mathbb Z$ there is an infinite number of integer triples $(a,b,c)$ which are solutions of $a^2+b^2+k=c^2$?
Yes, for any $k$, there are infinite many integer solutions.
For any odd number $\ell$, $k + (k+\ell)^2$ is odd. You can verify following 3 integers
$$(a,b,c) = \left( k+\ell, \frac{k + (k+\ell)^2 - 1}{2}, \frac{k + (k+\ell)^2+1}{2} \right)$$ satisfies $a^2 + b^2 + k = c^2$