Let $(a,b)$ be a pair of numbers that satisfy all conditions and let $k=\frac{(a+b)^2+4a}{ab}$. Consider the quadratic equation $$\frac{(x+b)^2+4x}{xb}=k \Rightarrow x^2-(kb-2b-4)x+b^2=0.$$One solution is $a$. By Vieta there is another solution $a'$ such that \begin{align*} a+a'&=(k-2)b-4, \\ aa'&=b^2. \end{align*}From the first equation we deduce that $a'$ is an integer. Let $d=\gcd(a,a')$. Since $d\mid b$ and $d\mid (k-2)b-4$ we know that $d\mid 4$. Using the fact that $b$ is odd we get $d=1$. Thus, $a$ and $a'$ are perfect squares.