This solution works iff $q$ is a prime power, so it's not a full solution, but I guess it's better than nothing, as it is completely elementary.
If $q$ is a prime power, then $q$ must fully divide either $s$ or $s-1$. We may WLOG assume it divides $s$ (since otherwise we may take $t=1-s$ as $s$). Then we have that, letting $s=mq$,
$$2(q-1)^2(q+1)=m(mq-1).$$
Taking $\bmod q$, we have that $m\equiv -2\bmod q$, so we may set $m=(x+1)q-2$, giving $s=-2q+(x+1)q^2$ and
$$(x^2+2x-1)q^2 - (4x+2) q + (5-x) = 0,$$
as noted in the question. From this, we have $x\equiv 5\bmod q$. Let $x=nq+5$. Then
$$s=-2q+(nq+6)q^2=nq^3+6q^2-2q=6q^2+q(nq^2-2).$$
We will prove the following lemma:
Lemma: If $q$ is a nonzero integer and
$$(2s-1)^2 = 8(q-1)^2q(q+1)+1,$$
then $|s|<3q^2.$
Proof: First, note that
$$3x^4-(x-1)^2x(x+1)=2x^4+x^3+x^2-x=(2x^2-x)(x^2+x+1)$$
is positive for all $x$ outside of $[0,1/2]$, and specifically is positive if $x$ is a nonzero integer. Thus,
$$(x-1)^2x(x+1)<3x^4.$$
$$(2s-1)^2=8(q-1)^2q(q+1)+1<24q^4+1\leq 25q^4.$$
So,
$$|2s-1|<5q^2.$$
$$|2s-1|+|1|<5q^2+1\leq 6q^2.$$
$$2|s|< 6q^2,$$
implying the result.
If $n=0$, then $s=-2q+6q^2$, and we have
$$(6q^2-2q)(6q^2-2q-1)=2(q-1)^2q(q+1).$$
This quartic has a triple root at $q=0$ and a single root at $q=11/17$, so we can safely ignore this case. Thus, $|n|\geq 1$. If $|q|\geq 11$, then $|nq|\geq 11$. So,
$$|nq^2|\geq 11|q|.$$
$$|nq^2|\geq 9|q|+2.$$
$$|nq^2-2| \geq 9|q|.$$
But
$$s=6q^2+q(nq^2-2).$$
So, since $|nq^2-2|\geq 9|q|$, we have that $|s|\geq 3q^2,$ which contradicts our lemma. Thus, we have reduced the problem to a finite case check on integers $-10\leq q\leq 10$, which can easily be done.
Best Answer
Determining which integers $n$ are a sum of three cubes is a very famous open problem:
$$a^3 + b^3 + c^3 = n, \quad a,b,c \in \mathbb{Z}.$$ Conjecturally, $n$ is a sum of three cubes iff $n \not \equiv 4,5 \bmod 9$.
Note that this is really a family of Diophantine equations, rather than a single Diophantine equation.