In the post Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the single positive integer solution $(x,y)=(18,7)$ is found using algebraic integers.
In one of the comments (Finding all solutions to $y^3 = x^2 + x + 1$ with $x,y$ integers larger than $1$, the OP indicated the method suggested (i.e., the “theory of elliptic curves”, torsion groups, etc.) was “far too advanced for [his] level of understanding”. This inspired me to try finding a totally elementary approach. I've made it to a certain stage, and wanted some advice on how to proceed.
Beginning with the original equation $$y^3 = x^2+x+1,\tag{1}$$ I added $x^3$ to both sides and factored, obtaining \begin{align} x^3+y^3 &= x^3+x^2+x+1 \\ (x+y)(x^2-xy+y^2) &= (x^2+1)(x+1).\tag{2} \end{align} Since we're looking for $x,y > 1$, and both factors on the right-hand side of (2) are positive, so are both on the left-hand side. Hence there exist positive integers $a,b,c,d$ such that \begin{align} x^2+1 &= ab, \\ x+1 &= cd, \\ x+y &= ac, \\ x^2-xy+y^2 &= bd. \end{align} Knowing the solution a priori, I'm now faced with trying to show that $(a,b,c,d)=(25,13,1,19)$. Using a few simple congruence and divisibility arguments, it can fairly easily be shown that $c=1$, and hence we have \begin{align} x^2+1 &= ab, \tag{3.1} \\ x+1 &= d, \tag{3.2} \\ x+y &= a, \tag{3.3} \\ x^2-xy+y^2 &= bd. \tag{3.4} \end{align} But now I'm running in circles. Evidently $(x,y)=(d-1,a-d+1)$, and I can prove other results like $(y-1)\mid x$ and $y \mid (b+1)$, but I can't seem to take it across the goal line. Any help would be greatly appreciated.
EDIT: From $a^2-bd=3xy$ and $d^2-ab=2x$, we have $$ (a-d)(a+d+b)=x(3y-2). \tag{4} $$ By the form of the left-hand side of (3.4) and the fact that $b,d$ are odd and relatively prime [because $x^2+1$ and $x+1$ are], we deduce $b\equiv d\equiv 1\!\pmod{6}$. Then (3.1) implies $a \equiv 1\!\pmod{6}$; in fact, by the form of (3.1), we have also $a \equiv b \equiv 1\!\pmod{12}$. In any case, $a+b+d \equiv 3\!\pmod{6}$, and (4) now implies $18 \mid x$.
EDIT: Another thread on the same question is https://mathoverflow.net/questions/56338/is-n-m-18-7-the-only-positive-solution-to-n2-n-1-m3.
So I've found an elementary and fairly easy method of proof that the only two positive integer solutions are $(x,y)=(0,1)$ and $(x,y)=(18,7)$, with the only two negative integer solutions being $(x,y)=(-1,1)$ and $(x,y)=(-19,7)$.
Using the logic as shown in the question, we quickly deduce that $(y-1) \mid x$, say $x = (y-1)w$ for integer $w$. Substituting into the original equation and solving for $y$ yields $$ y = \frac{w^2-1 \pm \sqrt{w^4-6w^2+4w-3}}{2}. \tag{$\star$} $$ Evidently, $w^4-6w^2+4w-3$ must be an integer square. Now observe that $$ (w^4-6w^2+4w-3)-(w^2-3)^2 = 4w-12 \tag{1} $$ and $$ (w^2-2)^2-(w^4-6w^2+4w-3) = 2w^2-4w+7. \tag{2} $$ The right-hand side of (2) is positive for all integer $w$, and the right-hand side of (1) is positive for $w>3$. So for $w > 3$ $$ (w^2-3)^2 < w^4-6w^2+4w-3 < (w^2-2)^2, $$ and the expression cannot be a square. For $w < 0$, we determine that $$ (w^2-4)^2 < w^4-6w^2+4w-3 < (w^2-3)^2, $$ and again it cannot be a square. Hence we need only consider $0 \le w \le 3$, and it is quickly found that the only solution is $w=3$. By substitution into $(\star)$, we find $y=1$ or $y=7$, as claimed.
p.s. Thanks to Will Jagy for the elegant bounds-based solution to the last step!