**This is a partial answer.**\
Using the group law of elliptic curves, we have,
$b_{10}=\frac{\left(n^{102}+133370 n^{100}+235431945 n^{98}-53960558412 n^{96}+\,\dots\,+607383986505 n^{6}+2125016730 n^{4}-948799 n^{2}+100\right) n}{100 n^{102}-948799 n^{100}+2125016730 n^{98}+607383986505 n^{96}+\,\dots\,-53960558412 n^{6}+235431945 n^{4}+133370 n^{2}+1}$
All other $b_{m}$ can also be generated. Given,
$$(a^3-b)(b^3-a) = y^2$$
---
**I. First family ($a=n$)**
Denote $$E_1=\{(U,V): V^2 = -U^4+n^3U^3+nU-n^4\}\cup O.$$
It is birationally equivalent to Weierstrass form
$$E_2=\{(X,Y): Y^{2}+\left(3 n^{2}-1\right)XY + \left(2 n^{6}-10 n^{4}+8 n^{2}\right)Y
=
X^{3}+\left(\frac{3}{4} n^{4}-\frac{9}{2} n^{2}-\frac{1}{4}\right) X^{2}+4 n^{2} \left(n -1\right)^{2} \left(n +1\right)^{2} X +3 n^{10}-24 n^{8}+38 n^{6}-16 n^{4}-n^{2}
\}\cup O$$
by,
$$\small{U =
\frac{-3 n^{7}+21 n^{5}+\left(-4 X -17\right) n^{3}+\left(4 X +2 Y -1\right) n}{2 Y}\\
V =
-\frac{\left(n +1\right) n \left(-3 n^{10}+\left(\frac{9 X}{8}+24\right) n^{8}+\left(-\frac{35 X}{2}+\frac{Y}{4}-38\right) n^{6}+\left(\frac{9}{4} X^{2}+\frac{191}{4} X -\frac{17}{4} Y +16\right) n^{4}+\left(-\frac{27}{2} X^{2}+\frac{1}{2} X -\frac{17}{4} Y +1\right) n^{2}+X^{3}-\frac{3 X^{2}}{4}+\frac{X}{8}+\frac{Y}{4}\right) \left(n -1\right)}{Y^{2}}\\
X =
\frac{n \left(3 U \,n^{4}-n^{5}-4 U \,n^{2}-2 V \,n^{2}+U +2 V +n \right)}{\left(U -n \right)^{2}}\\
Y =
-\frac{3n \left(n +1\right) \left(n -1\right)} {2 \left(U -n \right)^{3}} \left(-\frac{n^{6}}{3}+2 U \,n^{5}+\left(U^{2}-6\right) n^{4}+\frac{4 \left(5 U -2 V \right) n^{3}}{3}+\left(-6 U^{2}+1\right) n^{2}+2 \left(U +\frac{4 V}{3}\right) n -\frac{U^{2}}{3}\right)}$$
Let
$P=\left(3 n^{2}+1, -2 n^{6}-\frac{1}{2} n^{4}-5 n^{2}-\frac{1}{2}\right)\in E_2,$
then the $U$ corresponding to $[m-1]P$ is exactly $b_m$ where $m\geq 1$.
---
**II. Second family ($a=n^3$)**
Denote $$E_3=\{(U,V):V^{2} = n^{9} U^{3}-n^{12}-U^{4}+n^{3} U\}\cup O.$$
It is birationally equivalent to
$$E_4=\{(X,Y):Y^{2} =
X^{3}+\left(3 n^{10}-6 n^{2}\right) X^{2}+\left(3 n^{20}-15 n^{12}+12 n^{4}\right) X -9 n^{22}+18 n^{14}-9 n^{6}
\}\cup O$$
by $$\small{\left[U = \frac{n \left(3 n^{10}-3 n^{2}+X \right)}{X},\, V =
\frac{3 Y \,n^{11}-3 Y \,n^{3}}{X^{2}},\, X =
\frac{3 n^{11}-3 n^{3}}{U -n},\, Y =
\frac{3 V \,n^{11}-3 V \,n^{3}}{\left(U -n \right)^{2}}\right]}
$$
Let $Q=(n^{12}-n^{10}+n^{6}+2 n^{2}+1, n^{18}+n^{12}+n^{6}+1) \in E_4,$ then the $U$ corresponding to $[m]Q$ is exactly $b_m$ where $m\geq 1$. For example,
$b_1=\frac{\left(n^{6}+n^{4}-2 n^{2}+1\right) n}{n^{6}-2 n^{4}+n^{2}+1}$
$b_2=\frac{n \left(n^{12}+8 n^{10}+10 n^{6}-4 n^{2}+1\right)}{n^{12}-4 n^{10}+10 n^{6}+8 n^{2}+1}$
$b_3=\frac{\left(n^{30}+17 n^{28}-18 n^{26}+101 n^{24}-172 n^{22}+80 n^{20}+282 n^{18}-82 n^{16}-244 n^{14}+282 n^{12}-28 n^{10}-64 n^{8}+101 n^{6}+9 n^{4}-10 n^{2}+1\right) n}{n^{30}-10 n^{28}+9 n^{26}+101 n^{24}-64 n^{22}-28 n^{20}+282 n^{18}-244 n^{16}-82 n^{14}+282 n^{12}+80 n^{10}-172 n^{8}+101 n^{6}-18 n^{4}+17 n^{2}+1}
$
$b_4=\frac{n \left(n^{48}+32 n^{46}+552 n^{42}-1088 n^{40}-16 n^{38}+5820 n^{36}+8160 n^{34}+544 n^{32}+6552 n^{30}+18560 n^{28}-4080 n^{26}+23302 n^{24}+8160 n^{22}-9280 n^{20}+6552 n^{18}-1088 n^{16}-4080 n^{14}+5820 n^{12}+32 n^{10}+544 n^{8}+552 n^{6}-16 n^{2}+1\right)}{n^{48}-16 n^{46}+552 n^{42}+544 n^{40}+32 n^{38}+5820 n^{36}-4080 n^{34}-1088 n^{32}+6552 n^{30}-9280 n^{28}+8160 n^{26}+23302 n^{24}-4080 n^{22}+18560 n^{20}+6552 n^{18}+544 n^{16}+8160 n^{14}+5820 n^{12}-16 n^{10}-1088 n^{8}+552 n^{6}+32 n^{2}+1}
$
$b_5=\frac{n \left(n^{78}+49 n^{76}-50 n^{74}+2093 n^{72}-9092 n^{70}+7024 n^{68}+71486 n^{66}+146362 n^{64}-221372 n^{62}+120846 n^{60}+1989884 n^{58}-1997248 n^{56}+2411691 n^{54}+2513095 n^{52}-3945398 n^{50}+4925063 n^{48}+1374136 n^{46}-4812512 n^{44}+9246036 n^{42}+3689036 n^{40}-10556008 n^{38}+9246036 n^{36}+2487832 n^{34}-5926208 n^{32}+4925063 n^{30}+3430735 n^{28}-4863038 n^{26}+2411691 n^{24}+1056332 n^{22}-1063696 n^{20}+120846 n^{18}+102298 n^{16}-177308 n^{14}+71486 n^{12}-3476 n^{10}+1408 n^{8}+2093 n^{6}+25 n^{4}-26 n^{2}+1\right)}{n^{78}-26 n^{76}+25 n^{74}+2093 n^{72}+1408 n^{70}-3476 n^{68}+71486 n^{66}-177308 n^{64}+102298 n^{62}+120846 n^{60}-1063696 n^{58}+1056332 n^{56}+2411691 n^{54}-4863038 n^{52}+3430735 n^{50}+4925063 n^{48}-5926208 n^{46}+2487832 n^{44}+9246036 n^{42}-10556008 n^{40}+3689036 n^{38}+9246036 n^{36}-4812512 n^{34}+1374136 n^{32}+4925063 n^{30}-3945398 n^{28}+2513095 n^{26}+2411691 n^{24}-1997248 n^{22}+1989884 n^{20}+120846 n^{18}-221372 n^{16}+146362 n^{14}+71486 n^{12}+7024 n^{10}-9092 n^{8}+2093 n^{6}-50 n^{4}+49 n^{2}+1}
$
and so on.
---
**Note 1:**
In the first family, I think it is nice to denote $b_1=\frac{n\times \color{red}{1}}{1}.$
For the first family, the sum of the coefficients of $b_m$'s denominator is $2^{k(m)}$ where $k(m)=0, 4, 8, 16, 24, 36, 48, 64, 80, 100, 120, 144, 168, 196, 224, 256.$ (See https://oeis.org/A137932).
For the second family, the sum of the coefficients of $b_m$'s denominator is $2^{k(m)}$ where $k(m)=0, 4, 8, 16, 24, 36, 48, 64, 80, 100, 120, 144, 168, 196, 224.$ (See https://oeis.org/A137932).
That is $k(m)= 2 {\lfloor \frac{m^{2}}{2}\rfloor}.$
I have verified the first for $1\leq m \leq 16 $ and the second for $1\leq m \leq 15. $
---
**Note 2:**
For the first family, the degree of $b_m$'s denominator is $d(m)=2\alpha(m)$ where
$\alpha(m)=0, 3, 4, 9, 12, 19, 24, 33, 40, 51, 60, 73, 84, 99, 112.$ (See https://oeis.org/A097063).
For the second family, the degree of $b_m$'s denominator is $d(m)=6\beta(m)$ where
$\beta(m)=1, 2, 5, 8, 13, 18, 25, 32, 41, 50, 61, 72, 85, 98, 113.$ (See https://oeis.org/A000982).
It is worth noting,
$$\alpha(m)-\beta(m)=(-1)^m$$
I have verified this for $1\leq m \leq 15 $. I think it is true for all $m\geq 1$.
---
**Note 3: identities**
The discriminant of $E_2$ is
$$\Delta=-432 n^{20}+864 n^{12}-432 n^{4} =
-432 n^{4} \left(n -1\right)^{2} \left(n +1\right)^{2} \left(n^{2}+1\right)^{2} \left(n^{4}+1\right)^{2}
.$$
Let $\Delta=0$, we have
$$n=0,\pm1,\pm I,\pm\frac{\sqrt{2}}{2}(1+I),\pm\frac{\sqrt{2}}{2}(1-I)$$
where $I^2=-1.$ Some identities are related to these values.
In the following we denote $b_m(n)=n\frac{P_m(n)}{Q_m(n)}.$
For the first family, $P_m(0)=m^2$,$Q_m(0)=1$, $P_m(1)=Q_m(1)=2^{k(m)}$, and Sidharth Ghoshal also noticed that the values of $$\left|\frac{Q_{m+1}(I)}{Q_m(I)}\right|=12,144,144, 1728, 1728, 20736, 20736$$ for $2 \leq m\leq 8$ are all powers of $12$. A natural question may be asked: what are the values of $$P_m(n),Q_m(n)$$
where $n=I,\frac{\sqrt{2}}{2}(1+I)$? We have the following
**Identitiy 1.** $$P_m(I)=(-1)^{\frac{k(m+2)}{4}}12^{\frac{k(m)}{4}},$$
$$Q_m(I)=(-12)^{\frac{k(m)}{4}},$$
where $k(m)= 2 {\lfloor \frac{m^{2}}{2}\rfloor}.$
I have verified it for $1\leq m \leq 16.$
**Identitiy 2.** $$P_m\left(\frac{\sqrt{2}}{2}(1+I)\right)=
\begin{cases}
6^{\frac{m^2}{4}}I, & \text{if } m \equiv 0\pmod{4} \\
-6^{\frac{m^2}{4}}, & \text{if } m \equiv 2\pmod{4} \\
6^{\frac{m^2-1}{4}}(\lambda(m)+(\lambda(m)-1)I), & \text{if } m \equiv 1,7\pmod{8} \\
6^{\frac{m^2-1}{4}}(-\lambda(m)-(\lambda(m)-1)I), & \text{if } m \equiv 3,5\pmod{8}
\end{cases},$$
$$Q_m\left(\frac{\sqrt{2}}{2}(1+I)\right)=
\begin{cases}
6^{\frac{m^2}{4}}, & \text{if } m \equiv 0\pmod{4} \\
6^{\frac{m^2}{4}}I, & \text{if } m \equiv 2\pmod{4} \\
6^{\frac{m^2-1}{4}}(\lambda(m)-(\lambda(m)-1)I), & \text{if } m \equiv 1,7\pmod{8} \\
6^{\frac{m^2-1}{4}}(-\lambda(m)+(\lambda(m)-1)I), & \text{if } m \equiv 3,5\pmod{8}
\end{cases}
,$$
where $\lambda(m)= a_{\frac{m+1}{2}}=1, 5, 45, 441, 4361, 43165, 427285, 4229681,\cdots$(see https://oeis.org/A054318)
i.e, $$\lambda(m)=\frac{1}{2}+\frac{\left(\sqrt{3}+\sqrt{2}\right)^{m} \sqrt{3}}{12}+\frac{\left(\sqrt{3}-\sqrt{2}\right)^{m} \sqrt{3}}{12}.$$
I have verified it for $1\leq m \leq 16.$
For the second family, I think there are some similar identities.