<s>[INST] Trying to solve $x^2 – 11x = y^2$ knowing that $x$ and $y$ are integers. Once upon a time, I knew how to solve this equation. Yesterday, I tried (and failed) to solve it again. I remember the solutions, but I can't figure out how to find them…
 [/INST] You can factor the equation by completing the square and clearing denominators and you end up getting
$$
(2x - 11 - 2y)(2x - 11 + 2y) = 121
$$
After this you can consider all possible factorizations of $121$, so you'll have to solve some systems of linear equations.
Added in response to comments
To factor the original equation by completing the square and clearing denominators you can do the following.
$$
\begin{align}
x^2 - 11x = y^2 &\iff x^2 - 11x - y^2 = 0\\
&\iff (x^2 - 11x \quad \quad \quad) - y^2 = 0\\
&\iff \left( x^2 - 11x + \color{red}{ \left( \frac{11}{2} \right)^2}  -  \color{red}{ \left( \frac{11}{2} \right)^2} \right) - y^2 = 0\\
&\iff \left( x^2 - 11x +  \color{red}{ \left( \frac{11}{2} \right)^2} \right) - y^2 = \color{red}{ \left( \frac{11}{2} \right)^2}\\
&\iff \left( x - \frac{11}{2}  \right)^2 - y^2 = \frac{121}{4}\\
&\iff \left( \frac{2x - 11}{2}  \right)^2 - y^2 = \frac{121}{4}\\
&\iff \frac{1}{4} (2x - 11)^2 - y^2 = \frac{121}{4}\\
&\iff \hspace{-2.0cm}\underbrace{(2x - 11)^2 - 4y^2}_{\text{Difference of squares, factors as $A^2 - B^2 = (A-B)(A+B)$}} \hspace{-2.0cm}= 121\\
&\iff (\underbrace{2x - 11}_{A} -\underbrace{2y}_{B})(2x - 11 + 2y) = 121
\end{align}
$$
The method's name is just completing the square, and you should google for this if you want more information. In fact you can find some online videos where the method is explained, like this one. 
Finally, yes, all you have to do after factoring the equation is considering the pairs of equations you mentioned $(2x - 11 -2y, 2x -11 + 2y) = (\pm 11, \pm 11), (\pm 121, \pm 1), (\pm 1, \pm 121)$. Just be careful that for some of these you may not get solutions (I only solved a couple of them so I don't know if all of them give integer solutions).
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