The statement that $\gcd(a^3+6b^2:6480)=216$ means that you know $a^3+6b^2$ is not divisible by $5$, is divisible by $2^3$ but not $2^4$ and $3^3$ but not $3^4$.
This means that $a$ is divisible by $6$, and hence so must $b$ be too, so $a=6a_0,b=6b_0$ and $a_0^3+b_0^2$ is relatively prime to $30,$ and $11a_0+8b_0\mid 90$.
The general solution to $11x+8y=D$ is $x=3D-8k, y=-4D+11K$. If $a_0=3D-8k,b_0=-4D+11k$ then $$\begin{align}a_0^3 +b_0^2 &= (3D-8k)^3+(-4D+11k)^2 \\
&= 27D^3-27\cdot 8D^2k+9\cdot 8^2Dk^2-8^3k^3 + 16D^2-8\cdot 11Dk+11^2k^2 \\
&= 27D^3-D^2(27\cdot 8k-16)+D(9\cdot 8^2k^2-8\cdot 11k) +(11^2k^2-8^3k^3)\end{align}$$
Modulo $2$, this is $D^3+k^2$, so $D,k$ must be opposite parity.
Modulo $3$, this is $D^2-Dk+k^2+k$, and you want $D,k$ so that this is not divisible by $3$.
Modulo $5$, this means that $2D^3-D^2(k-1)+D(k^2-3k)+(k^2-2k^3)$ is not divisible by $5$.
A lot of tedious enumeration work will go into this for each $D$.
Case $D=30$ or $D=90$
For example, if $D=30$ or $D=90$, then the above reduces to $k$ being odd, $k^2+k$ not divisible by $3$ (so $k\equiv1 \pmod 3)$ and $k^2-2k^3$ not divisible by $5$, so $k\equiv 1,2,4\pmod 5$.
This means that when $D=30$ or $D=90$ then $k\equiv 1,7,19\pmod{30}$ and $(a,b)=(6(3D-8k),6(-4D+11k))$ is a solution.
One example would be $D=30, k=7$ which gives $(a,b)=(204,-258)$.
Case $D=1$
If $D=1$, then $k$ must be even, $k^2+1$ is not divisible by $3$ (true for any $k$) and $2-(k-1)+(k^2-3k)+(k^2-2k^3) = 3+k+2k^2-2k^3$ must not be divisible by $5$, so $k\equiv 0,1,2,4\pmod{5}$.
So when $D=1$, $k\equiv 0,2,4,6\pmod{10}$, and $(a,b)=(6(3-8k),6(-4+11k))$. At $k=0$, this gives $(a,b)=(18,-24)$.
Other cases
It gets messier for other values of $D$. Not sure if there is a better way than just enumerating $D\mid 90$ and then working it out in each case. We've already done $D=1,30,90$. There are $9$ other possible $D$.