If is $a$ is not a perfect square, then $d(a)$ is even, hence \[ a-b \equiv b^2 + 2 - d(a)a-b \equiv b^2-b \equiv 0 \pmod{2} \]If $a=k^2$ is a perfect square, then obviously $k=1$ fails. If $k>1$, then \[ b^2+2 = k^2 + d(k^2) \le k^2 + 2k - 1 < (k+1)^2 \]and \[ b^2+2 = k^2 + d(k^2) > k^2 +2 \]Hence, no $a, b$ satisfies this case.