Find all pairs of $(x, y ,z)$ such that $x + y =\sqrt{z^{2} + 2018}, .... $ Find all pairs of $(x,y,z)$, of real numbers, such that
$$ x + y = \sqrt{z^{2} + 2018} $$
$$ x + z = \sqrt{y^{2} + 2018} $$
$$ y + z = \sqrt{x^{2} + 2018} $$

An attempt : 
Squaring we get
$$ (x + y)^{2} = z^{2} + 2018 \implies  (x + y)^{2} - z^{2} = 2018 $$
$$ (x + z)^{2} = y^{2} + 2018 \implies  (x + z)^{2} - y^{2} = 2018 $$
$$ (y + z)^{2} = x^{2} + 2018 \implies  (z + y)^{2} - x^{2} = 2018 $$
which also means
$$ (x + y - z)(x + y + z) = 2018 $$
$$ (x + z - y)(x + y + z) = 2018 $$
$$ (z + y - x)(x + y + z) = 2018 $$
so
$$ \frac{2018}{x+y-z} =  \frac{2018}{x+z-y} =  \frac{2018}{y+z-x} $$
$$(x+y-z) = (x+z-y) = (y+z-x)$$
$$y-z = z-y \implies z = y $$
$$x-y= y-x \implies x=y $$
so my answer is
$$(x, y, z), \:\:\: x=y=z $$
but with the 3 initial equations, we must also have
$$ (x+y) = \sqrt{z^{2} + 2018} \implies 4x^{2} =  x^{2} + 2018 $$
or
$$ x^{2} = 2018/3 $$
so the solution is
$$(x, y, z), \:\:\: x=y=z = \sqrt{2018/3}  $$
Is this sufficient already? are there better techniques?

At the stage you have...
$$ (x + y - z)(x + y + z) = 2018 $$
$$ (x + z - y)(x + y + z) = 2018 $$
$$ (z + y - x)(x + y + z) = 2018 $$
This implies...
$$x+y-z=x+z-y=z+y-x=\dfrac{2018}{x+y+z}$$
...unless $x+y+z=0$. 
Notice that $x+y+z=0$ implies $x+y=-z$ and so $-z=\sqrt{z^2+2018}$ which means $z^2=z^2+2018$ and so $z=0$. Likewise, $x=y=0$ (contradiction since $(0,0,0)$ isn't a solution).
Thus $x+y-z=x+z-y=z+y-x$ and so you got that $x=y=z$. But then $x+y=\sqrt{z^2+2018}$ implies that $2z=\sqrt{z^2+2018}$ and so $4z^2=z^2+2018$ and so $x=y=z = \pm \sqrt{\dfrac{2018}{3}}$. But $x+y $ etc. are square roots (thus non negative). So there is only one solution.
