Find all ordered pairs $(x,y)$ of positive integers such that the expression $x^2+y^2+xy$ is a perfect square. My approach so far:
Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.
$\implies (x+y) ^2-xy=n^2$
$\implies (x+y) ^2-n^2=xy$
$\implies (x+y+n) (x+y-n) =xy$
Only one case is possible:
When $x+y+n=xy$ and $x+y-n=1$. On adding, we get: $2(x+y) =xy+1$. Since LHS is an even number so is RHS. That's why, $xy$ must be odd$\implies x$ and $y$ both are odds. So I checked for few odd numbers and found out that for $x=3$ and $y=5,\;x^2+y^2+xy$ becomes $49$ which is a perfect square.. I didn't dare to check for more odds as there could be many...
Recently I found out that for $x=5$ and $y=16$(an even number), $x^2+y^2+xy=361=19^2.$ (Surprising!!) 
So now I can say, I am stuck very badly.. All of my observations miserably failed...
Please suggest something! 

$x^2+yx+y^2-n^2=0 \implies \triangle = 4n^2-3y^2= d^2$. At this point I propose you to look up a article by L.J. Mordell or Kneser in internet and they have a formula for all the finite solutions that are bounded above by the products of the coefficients $(4,-3,-1)$ of the Diophantine equation above.
