Note that $(a,b)$ are coprime. Note that $b\equiv -a\pmod{a+b}$ yields $a+b\mid b^2-1$. Likewise, $a\equiv b\pmod{a-b}$ gives $a-b\mid b^2-1$. Equipped with this, let's choose $a,b$ such that (a) both $a+b$ and $a-b$ are even, and (b) $a+b,a-b\mid b^2-1$. For this, it suffices to ensure $a^2-3b^2=-2$ has infinitely many odd solutions with $b>2$. Notice that if $a^2=3b^2-2$ then $a>b\sqrt{3}-1$ necessarily, as otherwise $3b^2-2b\sqrt{3}+1>3b^2-2$, not possible for $b$ large. Any such odd solution would necessarily imply $a,b$ are coprime. Furthermore, $ab+1\equiv b^2-1\equiv \frac{a^2-b^2}{2} = (a+b)\frac{a-b}{2}\equiv 0\pmod{a+b}$, and $a-b\mid ab-1$ is established similarly. The infinitude of the solutions of the above equation is well-known, see e.g. here.