Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers? 
Prove or disprove that if $t$ is a positive integer, $$f(x,y)=\dfrac{x^2+y^2}{xy-t},$$ then $f(x,y)$ has only finitely many distinct positive integer values with $x,y$ positive integers. In other words, there exist $k\in\mathbb N$ such that if $n\gt k$ then $f(x,y)=n$ has no positive integer solutions.

This problem is a generalization of this famous problem.


*

*Below is the list of the set of $f(x,y)$ with $t\le 10$ (may be incomplete):


{t,{f(x,y)}}=
{1,{5}}
{2,{4,10}}
{3,{3,4,8,13,17}}
{4,{5,26}}
{5,{13,25,37}}
{6,{6,10,50}}
{7,{5,8,9,20,29,41,65}}
{8,{4,10,18,34,82}}
{9,{5,29,61,101}}
{10,{20,122}}
Thanks in advance!

October 7, 2015. This is with $$ \frac{x^2 + y^2}{xy - t} > 0, $$
which I believe to be the intent of the question. 
I proved finiteness, with an explicit bound that is not that bad.
This works. Note that the original question requires $xy> t.$ Otherwise we could have listed $x=1,y=1,t=2$ to get
$(x^2 + y^2)/ (xy-t) = -2.$ This was not done. So we are keeping $xy>t>0,$ in
$$ \frac{x^2 + y^2}{xy-t} = q. $$ 
We have the arc of the hyperbola
$$ x^2 - q x y + y^2 = -tq $$ in the first quadrant $x,y > 0$ that lies in the sector of the first quadrant defined by
$$ 2 x \leq q y $$ and
$$ 2 y \leq q x. $$ Note that the points of intersection of the two boundary lines with the hyperbola branch give the two points with the minimum values of $x$ and of $y.$
As noted in the other answer, if there are any integer solutions $(x,y)$ with $q$ also an integer, then there is at least one solution between the indicated Hurwitz lines. 
Next, we always have $q \geq 3.$ In $ x^2 - q x y + y^2 = -tq ,$ if $q=1$ the quadratic form on the left hand side is positive definite and can never equal the right hand side, which is negative. If $q=2$ the quadratic form on the left hand side is positive semi-definite ($(x-y)^2$) and can never equal the right hand side, which is negative. 
The key to finiteness was simply the size of $xy/t.$ We know already that $xy > t,$ that is $xy/t > 1.$ By Lagrange multipliers, the smallest value occurs when $x = y,$ at which point
$$ \frac{xy}{t} = \frac{q}{q-2} = 1 +  \frac{2}{q-2}. $$
Once again by Lagrange multipliers, the largest value of $xy/t$ within the Hurwitz region occurs at the boundary point where one of the lines meets the hyperbola.  One of them is at
$$ y = \left( \frac{2}{q} \right) x. $$ Plugging this into $ x^2 - q x y + y^2 = -tq $ gives a nice value for $x^2,$ then $ y^2 = \left( \frac{4}{q^2} \right) x^2 $ gives a nice value for $y^2.$ These turn out to be
$$  x^2 = \frac{q^3 t}{q^2 - 4}, \; \; \; y^2 = \frac{4 q t}{q^2 - 4}.  $$
Together 
$$  x^2 y^2 =  \frac{4 q^4 t^2}{(q^2 - 4)^2},  $$ and
$$  x y =  \frac{2 q^2 t}{q^2 - 4},  $$ or
$$  \frac{x y}{t} =  \frac{2 q^2 }{q^2 - 4} =   \frac{2 q^2 - 8 }{q^2 - 4} +  \frac{8 }{q^2 - 4} = 2  +  \frac{8 }{q^2 - 4} .  $$
This gives the maximum. Since $q \geq 3,$ 
$$  \frac{x y}{t} \leq  2  +  \frac{8 }{3^2 - 4} = \frac{18}{5} = 3.6 .  $$
Here we finally return to integers. We have $x \geq 1,$ which tells us that a Hurwitz fundamental solution always has
$$ y \leq  \frac{18}{5} t. $$ Once again, Lagrange multipliers tell us that $x^2 + y^2$ is maximized at the boundary point $x=1$ on the curve $xy= 18t/5,$ so
$$ x^2 + y^2 \leq 1 + \frac{324}{25} t^2.  $$ However, $xy - t \geq 1,$ meaning $q \leq x^2 + y^2.$ We then get finiteness from
$$  q  \leq 1 + \frac{324}{25} t^2.  $$
Computations, as above, suggest the stronger $q \leq t^2 + 2 t + 2.$ 
As far as rapid computer computations, for a fixed $t,$ we can demand $1 \leq x \leq 4 t$ because $18/5 < 4.$ For each $x,$ we can then demand $1 \leq y \leq x$ along with the very helpful $x y \leq 4 t.$ Having found an integer quotient $q,$ we then keep only those solutions with $2x \leq qy$ and $2y \leq qx.$
To repeat the good part: if there is any solution $(x,y)$ then there is at least one fundamental solution, that is with 
$$  \color{blue}{ 2x \leq qy}$$ and $$  \color{blue}{ 2y \leq qx}.$$
For such a fundamental solution, we have
$$ \color{blue}{ 1 + \frac{2}{q-2} \leq \frac{xy}{t} \leq  2 + \frac{8}{q^2-4} }. $$
Since $x^2 - qxy + y^2$ is positive (semi)-definite when $q = 1,2,$ we know that $q \geq 3$ always. Thus $q^2 - 4 \geq 5.$ As $\frac{8}{5} \leq 2,$ we get
$$  \color{blue}{xy \leq 4t}.  $$
Here is a graph for $t=1, q=5,$ showing the region where fundamental solutions must lie:

