$x^2=y^3+1$ has only 6 rational solutions. This is an exercise comment made by Dale Husemoller's Elliptic curve on Chpt 1, sect 3, Exercise 1.
$y^2=x^3+1$. It follows from (3.5) that this is all of group $E(Q)$ where $E(Q)$ denotes the rational points of $E$ over $Q$. 

(3.5) Let $E$ be an elliptic curve defined by $y^2=x^3+a$ with $a$ being 6-th power free. Then torsion $\operatorname{Tor}(E(Q))=Z_6$ if $a=1$. 

$\textbf{Q:}$ How does this follow that $E(Q)$ has no more rational points. In other words, why I do not have infinite order elements? I do not know whether he is assuming some other background information. The most probable one would be $Q(\zeta_3)$'s integral closure being PID which relates to class number information. 
Ref: D.Husemoller Elliptic Curves Chpt 1, Sec 3. 

I doubt that from  $ (3.5) $ one can deduce that the elliptic curve $E:\space y^2=x^3+1$ has no more than six rational points and I think there's a misprint in between. You need, I guess,  "other background information" as you have written. Prove  your curve has rank $0$ could be difficult without help of another strong enough property. Am I wrong in this point of view?
We have the following six rational points: $$A=(2,3),B=(0,1),C=(-1,0),-A=(2,-3),-B=(0,-1)$$ and the point at infinity taken as the zero of the group as usual. 
Since $\dfrac{3-1}{2-0}=\dfrac{0-1}{-1-0}$, lines $AB$ and $BC$ are the same so $A+B+C=0$ by definition of the group law and  obviously $-A-B-C=0$ 
 
We prove that $A$ is a generator of the torsion group $$\{A,B,C,-A,-B,O\}$$ using the formulas of sum $-(A+B)$ and $-2A$ (note the attached figure shows that $2A=B$).
The formulas for $y^2=x^3+ax+b$ with $a=0$ and $b=1$ are
$$►(x_a,y_a)+(x_b,y_b)=-(p^2-x_a-x_b,\space y_a+p(p^2-2x_a-x_b))$$ where $p=\dfrac{y_a-y_b}{x_a-x_b}$.
$$►2(x,y)=(x,y)+(x,y)=-(p^2-2x,-y+p(p^2-3x))$$ where $p=\dfrac{3x^2}{2y}$.
Then easy calculations give $$\begin{cases}A=A\\2A=B\\3A=B+A=C\\4A=C+A=-B\\5A=3A+B=C+B=-A\\6A=2C=0\end{cases}$$
Besides
 $2B=4A=-B$ and $3B=-B+B=0$ and $2C=0$ i.e. $A$ has order $6$,$\space B$ has order $3$ and $C$ has order $2$ and it is so  with $-A,-B$ and $-C$.
