We are given that
$$a^2 + b^2 = Q(a + b) + R.$$
Since $R < a + b$, we have
$$(a+b)^2 \le 2(a^2+b^2) < 2(Q+1)(a+b),$$
which implies $a + b < 2(Q + 1)$, and hence $R < 2(Q + 1)$. Thus, we have
$$Q^2 \le Q^2 + R \le Q^2 + 2Q + 1 = (Q + 1)^2.$$
Since $Q^2 + R = 2010$, we must have $Q = \lfloor\sqrt{2010}\rfloor = 44$, and so $R = 2010 - 44^2 = 74$.
It remains to solve the equation
$$a^2 + b^2 = 44(a + b) + 74.$$
This can be rewritten as
$$(a - 22)^2 + (b - 22)^2 = 1042.$$

Let $x = (a - 22)^2$ and $y = (b - 22)^2$. WLOG assume $x \ge y$. Then we have $521 \le x \le 1042$. So it suffices to check $23^2 \le x \le 32^2$. Also, $x$ and $y$ must be odd since $x^2 + y^2 \equiv 2 \pmod 4$. We compute the values as follows.

Among these values, only $y = 81$ is a perfect square. Therefore, we can only have $(a - 22, b - 22) = (\pm 31, \pm 9)$. Since $a, b \in \mathbb{Z}^+$, this implies $(a, b) = (53, 13), (53, 31)$. However, we need $a + b > R = 74$. Therefore, only $(a, b) = (53, 31)$ is possible. We check that when $(a, b) = (53, 31)$, we have
$$a^2 + b^2 = 3770 = 44(53 + 31) + 74.$$
Therefore, this is the only solution up to permutation.