Is 615 a special number?

First check whether $x,n$ has to be odd or even. The right hand side has to be even since it is $2^n.$ So L.H.S has to be even. Since odd+odd=even, $x$ is forced to be odd. Now if we put $x$ as odd, the L.H.S ends in the digits $0,4,6.$ So for $2^n$ to end in these digits, $n$ is forced to be even. Now keep $n=2k.$ $x^2+615=2^{2k}\Rightarrow 2^{2k}-x^2=615\Rightarrow (2^k+x)(2^k-x)=615.$ The factorization of $615= 41\times 5\times 3.$ So after some trial and error we find that $2^k+x=123$ and $2^k-x=5$ for $x,n$ to be positive integers. By simultaneous solving, we get $n=12$ and $x=59$ which is the only solution. So the sum of all possible values of $(x+n)=12+59=\boxed{71}.$

Same as the other solution but to discover that $n$ has to be even take the congruence $\mod 3$ . $x^2 \equiv 2^n - 615 \equiv (-1)^n \mod 3$ . We know that $-1$ is not a quadratic residue $\mod 3$ so $n$ must be even.