EDIT:I did mistake here....see below

First it's easy to see $(a+b)|b^2-1,(a-b)|b^2-1$ So $a^2-b^2|2(b^2-1)$ as $GCD (a,b)=1$ So we can take $3b^2=a^2+2$....clearly it satisfies all conditions. Now let $a_n=x_n+3y_n,b_n=x_n+y_n$ Where $x^2_n-3y^2_n=1$(as by pell's equation it has infinitely many solutions so clearly there are infinitely many $a_n,b_n$)

subham1729 wrote: $GCD (a,b)=1$ The problem didn't mention this. But the claim will remain true. Say $\gcd(a,b)=g$. Then $a+b|b^2-1\implies g(x+y)|g^2y^2-1\implies g|g^2y^2-1\implies g=1$.

Claim: The equation $x^2-3y^2=-2$ has infinitely many solutions over positive integers. Proof. Note that $(x,y)=(1,1)$ holds the equation. If $x^2-3y^2=-2$, then $(2x+3y)^2-3(x+2y)^2=4x^2+12xy+9y^2-3x^2-12xy-12y^2=x^2-3y^2=-2$ hence $(x,y) \mapsto (2x+3y,x+2y)$ satisfies.$\square$ We observe that $x,y$ must be odd by mod $4$. If we choose $a=x$ and $b=y$, we have $x^2+2x+1>x^2+2=3y^2$ thus, $x+1>y\sqrt3$ and $2(y^2-1)=x^2-y^2$ which implies $2|x+y\iff x-y|y^2-1\iff x-y|xy-1$ and $2|x-y\iff x+y|y^2-1\iff x+y|xy+1$ as desired.$\blacksquare$