Rather than discuss $\max b_{i}$, I'll discuss the equivalent question of bounding $\min a_{i}$. The ABC conjecture implies that for all $\epsilon > 0$, $\min a_{i} \gg (c^{2}+1)^{n/2 - 1 - \epsilon}$. This is because
if we have a solution to $a^{2} + b^{2} = (c^{2}+1)^{n}$ with $a \ll (c^{2} + 1)^{n/2 - 1 - \epsilon}$, set $A = a^{2}$, $B = b^{2}$ and $C = (c^{2}+1)^{n}$. Assume for simplicity that $\gcd(a,b) = 1$. (It doesn't change much if $\gcd(a,b) > 1$.) Then $C \ll {\rm rad}(ABC)^{1+\delta}$ for all $\delta > 0$. This gives
$$ (c^{2}+1)^{n} < (ab(c^{2}+1))^{1+\delta} \ll ((c^{2}+1)^{n/2 - 1 - \epsilon} (c^{2}+1)^{n/2} (c^{2}+1))^{1+\delta} = ((c^{2}+1)^{n-\epsilon})^{1+\delta}
$$
which is a contradiction if $\delta < \frac{\epsilon}{n-\epsilon}$.

For $n = 3$, it is possible to construct a sequence of integers $c$ which getse close to this bound. In particular, let
$$ c_{k} = \frac{(2 + \sqrt{3})^{k} - (2 - \sqrt{3})^{k}}{\sqrt{3}} \quad a_{k} = \frac{(2 + \sqrt{3})^{k} + (2 - \sqrt{3})^{k}}{2}. $$
It is easy to see that $a_{k}, c_{k} \in \mathbb{Z}$, $c_{k}$ is even, and a somewhat tedious calculation shows that
$$
(c_{k}^{2} + 1)^{3} = a_{k}^{2} + \left(c_{k}^{3} + \frac{3}{2} c_{k}\right)^{2}.
$$
In particular $\min a_{i} \leq a_{k} \approx \frac{\sqrt{3}}{2} c_{k}$.

One could hope to generalize this construction by finding $c_{k}^{2} + 1 = d_{k}^{2} + e_{k}^{2}$, and $\frac{e_{k}}{c_{k}} \approx \sin\left(\frac{\pi}{2n}\right)$. Letting $\frac{e_{k}}{\sqrt{d_{k}^{2} + e_{k}^{2}}} = \sin(\theta_{k})$, this makes
$$
(c_{k} + i)^{n} = (d_{k} + ie_{k})^{n} \approx (c_{k}^{2} + 1)^{n/2} \left(\cos(\theta_{k}) + i \sin(\theta_{k})\right)^{n} = (c^{2} + 1)^{n/2} \left(\cos(n \theta_{k}) + i \sin(n \theta_{k}\right))
$$
and $\sin(n \theta_{k}) \approx \sin(\pi/2) = 1$. This boils down to finding points on the hyperboloid $x^{2}+y^{2} = z^{2} + 1$ that lie close to the line $y = \sin\left(\frac{\pi}{2n}\right) z$.