I just want to point out that Don B. Zagier's one-paragraph solution to this notorious problem is more or less well-known:
Let $m$ be the quotient. We may suppose that $(a, b)$ is a minimal positive solution of the equation $(a^2 + b^2) = m(ab + 1)$
(i.e. one with the smallest value of $a + b$) for this value of $m$. Without loss of generality, suppose that $b ≥ a ≥ 0$, and set $b' = ma - b$. Then $a^2 = bb' + m$. If $b'$ is positive, then this equation implies $b' = (a^2 - m)/b < a^2/b < b$, and $(a, b')$ is a smaller positive solution of the equation of which $(a, b)$ was supposed to be the minimal solution. If $b'$ is negative, then $m = a^2 - bb' ≥ a^2 + b > b > ma ≥ m$, a contradiction. Hence $b' = 0$ and $m = a^2$.
Note: This solution is just the explicit result of applying reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + mxy + y^2$, which is the unique reduced quadratic form in its equivalence class.
(cf. http://www-groups.mcs.st-andrews.ac.uk/~john/Zagier/Solution1.3.html)
Maybe E. Atanassov's solution was similar to the above solution by Zagier.