System of diophantine equations $x^2+3y=u^2$, $y^2+3x=v^2$ 
Solve the following system of Diophantine equations(the unknowns are positive integers):
  $$
\left\{ 
\begin{array}{c}
x^2+3y=u^2 \\ 
y^2+3x=v^2
\end{array}
\right. 
$$  

I worked as follows:
subtract the two equations to get: $4x^2-4y^2-12(x-y)=9y^2-9x^2\ \implies\ ... (x-y)(13x+13y-12)=0\implies x=y\  or\  13x+13y-12=0$
The first equation has infinite answers and the second has none(since $gcd(13,13)$ does not divide $12$), am I right??

We can without (much) loss of generality assume that $y\le x$. Note that $x^2+3y$ is a perfect square  greater than $x^2$. 
Thus we have $x^2+3y\ge (x+1)^2$. But $x^2+3y\lt (x+2)^2$.  It follows that  $3y=2x+1$. 
Since $y^2+3x$ is a perfect square, so is $9y^2+27x$, that is, $4x^2+31x+1$.  This has to be a square, so $4x^2+31x+1$ is equal to one of $(2x+1)^2$ or $(2x+2)^2$ and so on up to $(2x+7)^2$, since $(2x+8)^2$ is clearly too big. 
The case $(2x+1)^2$ does not work, and neither does $(2x+2)^2$, nor $(2x+3)^2$.  The case $(2x+4)^2$ gives $x=1$. The case $(2x+5)^2$ does not work. The case $(2x+6)^2$ gives $x=5$, which does not work because $y$ is not an integer, and the case $(2x+7)^2$ gives $x=16$. 
We conclude that the solutions are $x=1,y=1$, $x=16, y=11$, (and $x=11,y=16$).
