<s>[INST] When $x^2+6xy+y^2$ a square number? Find all natural numbers $x$ and $y$ such that $x^2+6xy+y^2$ is a square number.
For example, $(x,y)=(2,3)$ or $(x,y)=(3,10)$.
Obviously, we can consider $gcd(x,y)=1$.
 [/INST] For non-trivial cases, $xy\ne0$ 
Let $x^2+6xy+y^2=(x+ky)^2$ where $k$ is any integer
$\iff y(6x+y)=y(2kx+k^2y)\implies6x+y=2kx+k^2y\iff x(6-2k)=y(k^2-1)$
So, $\dfrac x{k^2-1}=\dfrac y{6-2k}=m$(say an integer)
As $x,y>0$ if $m<0,$ we need $6-2k<0\iff k>3\ \ \ \ (1)$ and $k^2-1<0\iff -1<k<1\ \  \ \ (2)$ 
There can be no $k$ satisfying both $(1),(2)$
Similarly if $m>0,$ we need $6-2k>0\iff k<3$ and $k^2-1>0\iff k>1$ or $k<-1$
$k<3,k<-1\implies k<-1$
or $k<3,k>1\implies1<k<3\implies k=2$
So, there are infinitely natural values of $x,y$
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