Solve the equation in natural numbers $x^3+2x+1=y^2$ The Question

Solve the equation in natural numbers:
$x^3+2x+1=y^2,x,y\in N$
The solution is:
$x=1,y=2;x=8,y=23$

My Understanding
The given equation can be written as $x(x^2 +2)=(y-1)(y+1)$, as we're working in $\mathbb{N}$ we've the following cases:

*

*Case 1: $x=y-1$ and $x^2 +2 =y+1$ using these we get $y^2 -3y +2=(y-2)(y-1)=0$ therfore it follows that $(x,y)=(1,2)$ is the only solution in this case.

*Case 2: $x=y+1$, and $x^2+2 =y-1$ , using these we get $y^2 +y+4=0$. As the quadratic is irreducible over $\mathbb{Z}$ we can conclude the non existence of solutions in this case.

Therefore we can conclude that the only solution in $\mathbb{N}^{2}$ is $(x,y)=(1,2)$
What about $(8,23)$?
 A: Solving this problem using Elliptic Curves is probably not the best idea.
Examining the solutions to $E:y^2=x^3+2x+1$ over finite fields we see that $E(\mathbb{F}_{11})=16$ and $E(\mathbb{F}_{19})=27$. This mean that $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $16\cdot11^{r_1}$ for some $r_1$ and $|E(\mathbb{Q})_{\mathrm{tors}}|$ divides $27\cdot19^{r_2}$ for some $r_2$. The only solution is $|E(\mathbb{Q})_{\mathrm{tors}}|=1$, and so the torsion group is trivial.
As for the rank of the curve, the method of descent applied with $p=2$ shows that $E$ has rank $1$ (this can be done in Magma or Sage).
Combining both results, we see that $E(\mathbb{Q})$ is generated entirely by the free point $(0,1)$ on the curve. The issue is, there is no simple way to reduce this to just the solutions over $\mathbb{Z}$. It defiantly can be done, Magma or Sage can tell us that the the points $(0,\pm1), (1,\pm 2)$ and $(8,\pm 23)$ are the only solutions. If this question is being asked to high schoolers or undergraduates, this is not the way to go about it.
