Perfect Squares

If $n=1$ then the expression is always zero. If $n=4$ the expression simplifies to $(x^2+xy+y^2)^2.$

Now let $x=-1,y=-1,z=2.$ The expression becomes $2^{n-1} + (-1)^n.$ Clearly this is not a square for $n=2,3,$ so we can restrict our attention to $n \ge 5.$

If $n$ is odd then we get $a^2-1 = b^2$ where $a = 2^{(n-1)/2}$ and $b$ is a positive integer, so $(a-b)(a+b) =1$ which is clearly impossible for positive integers $a,b.$

If $n$ is even then we get $2^{n-1} + 1 = b^2$ for some positive integer $b,$ so $(b-1)(b+1) = 2^{n-1}.$ So both $b-1$ and $b+1$ are powers of $2,$ but their gcd is at most $2,$ so one of them is $\le 2.$ So $b \le 3.$ But this is impossible for $n \ge 5.$

So there are no other admissible values of $n,$ so the answer is $1+4=\fbox{5}.$