Here's a start. If you notice, $\gcd(a,a^2+2)=\gcd(a,2)$, so
by the unique factorization theorem,
$$
\eqalign{
& 3a(a^2+2)=z^3 \\
& \text{is a perfect cube}
}
~\iff~
\matrix{
a &=& a_1^3\cdot2^{a_2}\cdot3^{a_3} \\
a^2+2 &=& b_1^3\cdot2^{b_2}\cdot3^{b_3}
}
\quad
\text{with}
\quad
\eqalign{
& 2,3\not\mid a_1,b_1 \\
& a_2+b_2\equiv0\pmod3 \\
& a_3+b_3\equiv2\pmod3,
}
$$
with all $a_i,b_i$ nonnegative except for
$a_1$, which has the same sign as $a$.
This has known solutions
$$a=0,\quad\text{for which}\quad
\mathbb{a}
=\left[\matrix{a_1\cr a_2\cr a_3}\right]
=\left[\matrix{0\cr0\cr0}\right],\quad
\mathbb{b}
=\left[\matrix{b_1\cr b_2\cr b_3}\right]
=\left[\matrix{1\cr1\cr0}\right],
\tag{solution 0}
$$
and
$$a=\pm4,\quad\text{for which}\quad
\mathbb{a}=\left[\matrix{\pm1\cr2\cr0}\right],\quad
\mathbb{b}=\left[\matrix{1\cr1\cr2}\right].
\tag{solution 1}
$$
But then for $a\ne0$, assume without loss of generality that $a>0$
(so $a_1,b_1\ge1$), and note that,
looking at the powers of two and three in its prime factorization,
$$
\eqalign{
b_1^3\cdot2^{b_2}\cdot3^{b_3}
& = a^2+2
= a_1^6\cdot2^{2a_2}\cdot3^{2a_3}+2 \\
& = 2 \left( a_1^6\cdot2^{2a_2-1}\cdot3^{2a_3} + 1\right)
}
$$
implies that either $a_2\ge1=b_2$ (case 2.1) or $a_2=b_2=0$ (case 2.0),
and by similar reasoning, either $a_3>0=b_3$ (case 3.1),
or else $a_3=0$ and $b_3\equiv2\pmod3$, which I will call case 3.0.
This leaves us with four combinations to check,
and hopefully either rule out or simplify:
$$
\matrix{
\\\text{case }0=2.0\wedge3.0:&\quad a_2=a_3=b_2=0 ,\quad b_3\equiv2\pmod3
\\\text{case }1=2.1\wedge3.0:&\quad a_2\ge1=b_2,\quad a_3=0,\quad b_3\equiv2\pmod3
\\\text{case }2=2.0\wedge3.1:&\quad a_3>0=a_2=b_2=b_3
\\\text{case }3=2.1\wedge3.1:&\quad a_2,a_3\ge1=b_2,\quad b_3=0.
}
$$
These four cases comprise all nontrivial
(i.e. other than solution 0), (WLOG) postivie solutions,
and we hope to deduce that only case 1 with solution 1 is viable.
Case 0 simplifies to
$$
b_1^3 \cdot 3^{b_3} = a_1^6+2
\qquad\text{for}\qquad
a_1,b_1\not\equiv0\pmod{2,3}
\quad\text{and}\quad
b_3\equiv2\pmod3.
$$
Since $b_3\ge2$,
we have $a_1^6\equiv-2\pmod9$, which has no solutions $a_1$
(sixth powers are always $0$ or $1$ modulo $9$), ruling this case out.
Case 2 simplifies to
$$
b_1^3 = a_1^6 \cdot 3^{2a_3} + 2
\qquad\text{for}\qquad
a_1,b_1\not\equiv0\pmod{2,3}
\quad\text{and}\quad
a_3 > 0,
$$
so that $b_1^3\equiv2\pmod9$, which we can also rule out,
since cubes are always $0$ or $\pm1$ modulo $9$.
Case 3 simplifies to
$$
2b_1^3 = 2^{2a_2 } 3^{2a_3} a_1^6 + 2
= 2\left( 2^{2a_2-1} 3^{2a_3} a_1^6 + 1 \right)
\qquad\text{for}\qquad
a_1,b_1\not\equiv0\pmod{2,3}
\quad\text{and}\quad
a_2,a_3 > 1,
$$
so that $b_1^3\equiv1\pmod9\implies b_1$ is even,
a contradiction. Lastly, case 1 simplifies to
$$
3^{b_3} \cdot b_1^3 = 2^{2a_2-1} a_1^6 + 1
\qquad\text{for}\qquad
a_1,b_1\not\equiv0\pmod{2,3},
\quad a_2 \ge 1
\quad\text{and}\quad
b_3\equiv2\pmod3.
$$
so that $b_3\ge2\implies$ $2^{2a_2-1} a_1^6\equiv-1\pmod9$
$\implies a_1^6 \equiv-2^{1-2a_2} \equiv 2^{-2(a_2+1)} \pmod 9$
since $2$ is a primitive root modulo $9$,
i.e. $\text{ord}_9(2)=6=\phi(9)$, and $2^{\pm3}\equiv-1\pmod9$.
But then $a_1\equiv2^{a_0}\pmod9$
for some $a_0$ with $-2(a_2+1)\equiv6a_0\equiv0\pmod6$,
so that $a_2+1\equiv3~\implies~a_2\equiv2\pmod6$.
In particular, we see that $a_2\ge2$.
Well anyway, that's a start.
If we could show that $a_2,b_3\le2$,
then we could reduce case 1 to $9y^3-8x^6=1$,
and our goal would be to show that
its only positive solution is $(1,1)$,
and a start at that might be to note
that $9y^3\equiv8x^6+1\equiv9\pmod{72}$.