Cube & Square Difference

First of all check the parity of an integral solution. Suppose that $y$ is odd, this implies that $x$ is even but

$x^3=y^2+11\qquad x=2x',y=2y'+1$

$\Longrightarrow 8x'^3=4y'^2+4y'+12$

$2x'^3=y'(y'+1)+3$

But the $LHS$ is even while the $RHS$ is odd, so our assumption was false and a solution (if there exist any) must have odd $x$ and even $y$

Now, since $11$ is a Heegner number , we can exploit the fact that the ring $\mathbb{Z}\left[\frac{1}{2}+\frac{\sqrt{-11}}{2}\right]=\left\{m+n\left(\frac{1}{2}+\frac{\sqrt{-11}}{2}\right) \mid m,n\in\mathbb{Z}\right\}$ is a UFD

Factor the $RHS$

$x^3=(y+\sqrt{-11})(y-\sqrt{-11})$

We wish to prove that this two factors are coprime in the ring

Define the norm function as

$N(m+n\sqrt{d})=m^2-n^2d$

It's easy to prove that it has the property

$\alpha\mid\beta \Longrightarrow N(\alpha)\mid N(\beta)$

We are now going to consider a $\delta$ such that

$\delta\mid y+\sqrt{-11}\qquad\delta\mid y-\sqrt{-11}$

It follows that

$N(\delta)\mid y^2+11$

And $y^2+11$ is odd so $N(\delta)$ must be odd too. In addition

$\delta\mid (y+\sqrt{-11})-(y-\sqrt{-11})\equiv \delta\mid 2\sqrt{-11}$

We now know that $N(\delta)$ is odd and divides $44$ , so it must be $11$ or $1$

If $N(\delta)=11$ then both $x$ and $y$ would be divisible by $11$ , implying that

$1331x''^3=121y''^2+11$

But $v_{11}(LHS)\geq 3$ while $v_{11}(RHS)=1$

Hence $N(\delta)=1$ and $\delta$ is a unit, so $y+\sqrt{-11}$ and $y-\sqrt{-11}$ are coprime

We have two coprime numbers whose product is a cube, so they must be two cubes themselves, up to some units

We can avoid the last restrinction noting that the units in the ring are $\pm 1$ which are both cubes

Now what is left are some calculations and casework

$y+\sqrt{-11}=\left[m+n\left(\frac{1}{2}+\frac{\sqrt{-11}}{2}\right)\right]^3$

$8y+8\sqrt{-11}=\left[2m+n\left(1+\sqrt{-11}\right)\right]^3$

$8y+8\sqrt{-11}=8m^3+12m^2n(1+\sqrt{-11})+6mn^2(1+\sqrt{-11})^2+n^3(1+\sqrt{-11})^3$

$8y+8\sqrt{-11}=8m^3+12m^2n-60mn^2-32n^3+(12m^2n+12mn^2-8n^3)\sqrt{-11}$

Equating coefficients and simplifying

$\begin{cases} 2y=2m^3+3m^2n-15mn^2-8n^3\\ 2=3m^2n+3mn^2-2n^3 \end{cases}$

From the second equation we get that $n\mid 2$ , so $n=\pm 1,\pm 2$

• $n=1$ yelds no integer values for $m$

• $n=-1$ gives $m=0$ or $m=1$ , which yelds $y=\pm 4$

• $n=2$ gives $m=1$ or $m=-3$ , which yelds $y=\pm 58$

• $n=-2$ yelds no integer values for $m$

To conclude, the solutions to the equation are $(3;\pm 4), (15;\pm 58)$ and the sum of all the distinct values of $x$ is $15+3=\boxed{18}$

I would be somewhat surprised if there were an elementary way to solve this problem. I used MAGMA:

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E := EllipticCurve([0,-11]);
E;
IntegralPoints(E);

with output

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Elliptic Curve defined by y^2 = x^3 - 11 over Rational Field
[ (3 : -4 : 1), (15 : 58 : 1) ]
[ <(3 : -4 : 1), 1>, <(15 : 58 : 1), 1> ]

Apparently this gives you all the integral points up to the sign of $y.$ So the answer is $3+15 = \fbox{18}.$

By the way, this particular curve has infinitely many rational points; MAGMA says it has rank 2 (and no nontrivial torsion).