The given equality can be written more conveniently as:
$$(a - b)(1 + 4(a + b)) = b^2.$$
It suffices to show that the bracketed expressions are coprime, since that implies that $a-b$ and $1+4(a+b)$ are squares of positive integers.
Denote by $d$ the greatest common divisor of $a-b$ and $1+4(a+b)$. We have
$$d = M(a - b, 1 + 4(a + b)) = M(a - b, 1 + 4(a + b) - 4(a - b)) = M(a - b, 1 + 8b).$$
From this, it follows that $d$ divides $1+8b$.
From the initial equality, we have that $d$ also divides $b^2$.
Since $M(1+8b, b^2) = 1$, we conclude that $d = 1$, i.e. $a-b$ and $1+4(a+b)$ are coprime, which completes the proof.