The numbers' escaping, through this square that is gaping!

Given the equation, $(a^{2}(b^{2} + 2) + 2ab + 1 )^{2} = 6(a^{2} b + a)^{2}$ . Therefore, $((ab+1)^{2}+2a^{2})= 6a^{2}(ab+1)^{2}$ => $\left(\dfrac{(ab+1)^{2}+2a^{2}}{a(ab+1)}\right)^{2}=6$ => $\left( \dfrac{ab+1}{a}+\dfrac{2a}{ab+1}\right)^{2}=6$ .

Let us now make a substitution $\left(\dfrac{ab+1}{a}\right)=x$ . So our equation becomes: $\left(x+\dfrac{2}{x}\right)^{2}=6$ => $x^{2}+ \dfrac{4}{x^{2}}-2=0 => x^{4}-2x^{2}+4=0$ (Since $x=\dfrac{ab+1}{a}\neq 0)$ .

Clearly the discriminant for this bi-quadratic equation is $\text{negative}$ . Therefore, $x^{2}$ and further $x$ satisfy non-real i.e imaginary values. Since $x$ is an algebraic combination of positive reals $a$ and $b$ , it can never yield imaginary values. $\left(\dfrac{ab+1}{a}\right)$ comes out to be equal to $\sqrt{1 \pm i\sqrt{3}}$

$\therefore$ Number of ordered pairs of $(a,b)$ = $0 :)$

$\text{Alternate solution:}$ Since $a,b$ are positive integers we can conveniently introduce AM-GM inequality for x: $x \geq 2\sqrt{\dfrac{ab+1}{a}\times\dfrac{2a}{ab+1}}= 2\sqrt{2}$ $\therefore x^{2}\geq 8$ Since the minimum value of $x^2$ is 8 it can never attain the value of 6.

$\therefore$ Number of ordered pairs of $(a,b)$ = $0 :)$

$(a^2(b^2+2)+2ab+1)^2= 6(a^{2}b+a)^2 \\ \implies (a^2 b^2 + 2ab+1+2a^2)^2=6a^2(ab+1)^2 \\ \implies ((ab+1)^2+2a^2)^2=6a^2(ab+1)^2\\$

Put x=ab+1 $\implies (x^2+2a^2)^2 = 6a^{2}x^{2}\\$ $\implies x^4-2a^{2}x^{2}+4a^4=0\\$ $\implies (x^2-a^2)^2+3a^4=0\\$ $\implies ((ab+1)^{2}-a^2)^2+(3a^2)^2=0\\$

Now sum of squares of two real numbers is zero means that they both must be equal to zero

$\implies 3a^2=((ab+1)^{2}-a^2)=0 \\ \implies a=0\\ \implies \text{the equation} (ab+1)^2=a^2 \text{becomes 1=0 which is a contradiction}$

So we get $\textbf{NO REAL}$ solutions.