I think famous Vieta-jumping might help!

Of course, standard Vieta Jumping, but I don't think I've ever seen this problem before - strange, considering it is quite old. Anyway, let $a^2+b^2+3=kab$. After letting $(A,B)$ in $\mathbb{N}^2$ be the pair minimising $A+B$ over all solutions to the equation $a^2+b^2+3=kab$, and considering the quadratic $v^2-kvB+B^2+3=0$ one sees that $A+u=kB$ and $Au=B^2+3$ where $u$ is the second root to the equation. Clearly $u$ is both positive and an integer, so $(u,B)$ is another solution to the equation $a^2+b^2+3=kab$. Assume $A>B\ge 2$. Then $A^2>B^2+3=Au$ i.e. $A>u$ but then $B+u$ would be minimal over all solutions $(a,b)$, of course a contradiction. So either $A=B$ or $B=1$. If $A=B$ then $A^2+B^2+3=kAB\implies (k-2)A^2=3$ but $3$ is square free so $A=1$ and $k=5$. If $B=1$ then $A^2+4=kA$. So $A\mid 4$ i.e. $A=1,2,4$ (note $LHS>0$). These lead to the values $5,4,5$ for $k$ respectively. So overall $k=4$ or $k=5$. And yes I admit I didn't read the question - I only realise now the question asks for the solutions $(a,b)$ rather than the values of $\frac{a^2+b^2+3}{ab}\in\mathbb{Z}$. I'm wondering - is it possible to find all solutions? One can easily find infinite families of solutions using the relations above but I'm struggling to see why there cannot exist another solution skipped over by these sets of solutions...

Just to be related with Vieta Jumping: See this. Download the attached file in that post, there is a very similar problems to this one ($a^2+b^2+1$ instead of $a^2+b^2+3$).

It's Pen A 89