Write $gcd(a,b)=d$, and $a=dx$, $b=dy$. Taking modulo $d$ we get $d\mid x+y$, and modulo $x+y$ we get $x+y\mid d$, which implies that $d=(x+y)$. Now, substituting into the original equation, and taking modulo $(x+y)^2$ we get that $(x+y)^2\mid xy-1 \implies x=y=1\implies a=b=2 \implies l=1$, so we're done.