the best behaved indefinite ternary quadratic form is $y^2 - zx.$ It takes little to prove that, demanding $x>0$ and $\gcd(x,y,z) = 1,$
$$ x = u^2, \; \; y=uv, \; \; z = v^2 $$
But then all solutions to your $bc-ca+ab=0$ come from
$$ a=x+y, \; \; b = y, \; \; c = y+z . $$
You wanted $a-b.$
Here
$$ a-b = x+y-y = x = u^2 $$
Alright, let me work on the reverse direction a bit
Not bad: In $y^2 - zx = 0,$ take $x=a-b, y = a, z=a+c.$ So that, demanding $x>0,$ it is a square. We have $\gcd(x,y,z) = \gcd(a,b,c)$ and a little effort tells us this is $1$
ADDED. There is already an answer that refers to unique factorization; here let me talk about $zx=y^2$ using just gcd. The extra demands are that $x >0$ and $\gcd(x,y,z) = 1.$ We begin with $g=gcd(z,y),$ so that $z=gs, y=gt,$ and $\gcd(s,t)=1.$
We reach $gsx = g^2 t^2,$ or $sx = g t^2.$ As $\gcd(s,t)=1,$ we know that $s|g$ so that we may write $g = su.$ This leads to $sx= sut^2$ and $x=ut^2,$ so that $u>0.$
Next we combine to get $y=stu$ and $z = us^2$
We have reached $$ x=u t^2, \; \; y = stu, \; \; z = u s^2 $$
However, as $u$ divides all three and $\gcd(x,y,z) = 1,$ we know $u=1$ Finally
$$ x= t^2, \; \; y = st, \; \; z = s^2 $$