I claim all such pairs are $(a,b)=(b+1,b)$ where $b\in\mathbb{N}$ is arbitrary. Let $d={\rm gcd}(a,b)$ with $a=da_1$ and $b=db_1$ to arrive at $d+3da_1b_1 = d^3(a_1^3-b_1^3)$, that is \[ 1+3a_1b_1 = d^2(a_1-b_1)(a_1^2+a_1b_1+b_1^2). \]Now, it is clear that $a_1\ge b_1+1$. Moreover, $a_1^2+a_1b_1+b_1^2\ge 3a_1b_1$, but note that the equality does not hold since $a_1>b_1$, so that $a_1^2+a_1b_1+b_1^2\ge 3a_1b_1+1$. Combining these facts, we find that $d^2(a_1-b_1)=1$, and working backwards yields $(b+1,b)$ with $b$ arbitrary as the only such family.