Find all possible integers $n$ such that $\sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer. 
Find all possible integers $n$ such that $m = \sqrt{n + 2} + \sqrt{n + \sqrt{n + 2}}$ is an integer.

Guess what? This problem is adapted from a recent competition. There have been a solution below for you to check out. I am aware of the fact that there are other solutions that are more practical and suitable in test setting.

A shorter approach:
Generally for numbers $m=\sqrt{n+a}+\sqrt{n+\sqrt{n+a}}$ we must have:
$n+a=b^2$
$b^2+b-a=c^2$
If this system of equations, with given a, has integer solutions for n then m is integer.For example $a=3$; $m=\sqrt{n+3}+\sqrt{n+\sqrt{n+3}}$ we have:
$n+3=b^2$ 
$b^2+b-3=c^2$
It can be seen that with $c=3$ we have:
$b^2+b-3=3^2=9$; or $b^2+b-12=0$ which gives $b=3$ and $b=-4$. With $b=3$ we get $n=6$ and $m=6$, with $b=-4$ we get $n+3=16$ or $n=13$ which gives $m=4+\sqrt{17}$  which is not integer. 
In question $a=2$ and  we can search and see that reasonable value of c can only be zero. Other values give irrational numbers for one of terms $\sqrt{n+2}$ or $\sqrt{n+\sqrt{n+2}}$. Hence we have:  
$$b^2+b-2=0$$
Which gives:
$b=1$; $a= 1$; $n=-1$; $m= 1$
$b=-2$;$a= 4$; $n=2$; $m=4$
