Let $u = x+4$ and $v = x-4$ . Then we have:

$\begin{aligned} (x+y)^2 & = (x+4)(x-4) \\ (u+v)^2 & = uv \\ u^2 + uv + v^2 & = 0 \\ \Rightarrow u & = \frac{-v \pm \sqrt{v^2-4v^2}}{2} \\ & = \frac{-v \pm \sqrt{\color{#3D99F6}{-3v^2}}}{2} \end{aligned}$

We note that $\color{#3D99F6}{-3v^2} \le 0$ therefore, $u$ is only real when $\color{#3D99F6}{-3v^2} = 0$ , that is $v = 0$ and $u=0$ and that $x = -4$ and $y = 4$ . Therefore, there is only $\boxed{1}$ pair of $(x,y)$ that satisfies the equation.

I have two ways to solve this problem

1

We set $a=x+4$ ; $b=y-4$ , then the equation becomes $(a+b)^2=ab$ We have this inequality for all reals $(a+b)^2\geq 4ab$ $\Leftrightarrow 3(a+b)^2\leq 0$ $\Rightarrow a=-b\Leftrightarrow x=-y$ So now we have $(x+4)(-x-4)=0\Leftrightarrow x=-4$ So there are one pair $(x,y)=(-4;4)$

2

We set $y=tx$ and the equation becomes $(t+1)^2x^2=(x+4)(tx-4)$ $\Leftrightarrow [(t+1)^2-t]x^2-4(t-1)x+16=0$ We see that $(t+1)^2\neq t$ so $x$ exist when $\Delta'\geq 0$ $\Leftrightarrow 4(t-1)^2-16(t+1)^2+16t\geq 0$ $\Leftrightarrow 12t^2+24t+12\leq 0$ $\Leftrightarrow t=-1$ With $t=-1$ we get $x=-4\Rightarrow y=4$ So there's only one pair $(x,y)$ satisfy the equation

Expanding and moving everything to one side gives us ${ x }^{ 2 }+(y+4)x+{ y }^{ 2 }-4y+16=0.$

To eliminate the $xy$ term we let $z=x+\frac { 1 }{ 2 } y$ and replace $x$ with $z-\frac { 1 }{ 2 } y$ . The equation above becomes ${ z }^{ 2 }+4(z-\frac { 1 }{ 2 } y)+\frac { 3 }{ 4 } { y }^{ 2 }-4y+16=0.$

However, ${ z }^{ 2 }+4(z-\frac { 1 }{ 2 } y)+\frac { 3 }{ 4 } { y }^{ 2 }-4y+16 = { (z+2) }^{ 2 }+\frac { 3 }{ 4 } { y }^{ 2 }-6y+12$

$={ (z+2) }^{ 2 }+\frac { 3 }{ 4 } ({ y }^{ 2 }-8y+16)$

$={ (z+2 })^{ 2 }+\frac { 3 }{ 4 } { (y+4) }^{ 2 }.$

So the only real solution is when $z=-2$ and $y=4$ ; i.e. $x=-4$ and $y=4.$