How to solve the problem on number theory Find the number of positive integer pairs $x,y$ such that 
$$xy+\dfrac{(x^3+y^3)}3=2007.$$
I solved the question by using factorization and further checking possible values of $x$ and $y$. But it was very lengthy as I had to check many cases for $x$ and $y$. Is there any possible other method?

I show here that the only correct answer is, as it maintain  @fleablood and @farruhota in their answers.
$$\frac{x^3+y^3}{3}=\frac{(x+y)^3}{3}-xy(x+y)\Rightarrow xy+\frac{(x+y)^3}{3}-xy(x+y)=2007$$ It follows $x+y\equiv 0\pmod3$.
The cubic $$xy+\frac{x^3+y^3}{3}=2007$$ is symmetric respect to the diagonal $y=x$  then if the line $x + y = 3a$ cuts the cubic at the point $(x, y)$ it also cuts it at the point $(y, x)$. The system
$$\begin{cases}x+y=3a\\xy+\dfrac{x^3+y^3}{3}=2007\end{cases}$$ has a solving equation of the second degree whose solution $(x,y)$ (and the symmetric $(y,x))$ is$$\begin{cases}x=\dfrac{3(-3a^2+a+X)}{2-6a}\\y=\dfrac{3(3a^2-+Xa+X)}{2-6a}\end{cases}\quad\text{ where } X=\sqrt{-3a^4-2a^3+a^2+2676a-892}$$ It follows that there are quite few values of $a$ giving real roots. In the attached  graph of the cubic, only the black lines $x + y = 21,24,27$ (corresponding to $a=7,8,9$) cut the curve in real positive points. For $a\le6$ and $a\ge10$ does not. For the line $x+y=18$ the red points of intersection are not in the first quadrant (approximately $(18.204,-0.204)$).
Finally we have just three candidates (and the symmetric) produced by the lines $x+y=21,24,27$ but only the line $x+y=21$  agrees with the requirement of positive integers for the intersection.  In this line we have marked all the points of integer coordinates  of the first quadrant (in red those of intersection with the cubic). These three candidates are
$$(18,3)\quad\text{ and approximately }\quad(17.56,6.44),(16.162,10.838).$$
Thus the only solutions are given by $\{18,3\}$.

