Please correct me if I am wrong. We get that $(a^2+b^2)\mid a^3-b^3=(a-b)(a^2+ab+b^2)$, thus if $\gcd(a,b)=d$, then $$(a_1^2+b_1^2)\mid d(a_1-b_1)(a_1^2+a_1b_1+b_1^2).$$From here, since $\gcd(a_1,b_1)=1$, obviously, $(a_1^2+b_1^2)\mid d$, but also we know that $d\mid a,b$, thus $$a_1d \ge d\ge a_1^2+b_1^2\implies a_1(d-a_1)\ge b_1^2\ge 0,$$thus $d\ge a_1$ and similarly $d\ge b_1$. Suppose $d=1$, thus $a_1=b_1=d=1=a=b=1$, which is indeed solution. Suppose $d>1$, thus $d^2(a_1^2+b_1^2)\mid a^3+1^3=(a_1d+1)(d^2a_1^2-da_1+1)$. Contradiction! $\boxed{\text{Answer. } a=b=1}$