Legend $2^{x}$

We notice that $1+2^x+2^{2x+1}$ is not an integer if $x<-1$ and equals $2$ and $4$ when $x= -1 \text{ and } 0$ respectively.

Thus, the only solution is when $x\le 0$ is $(x,y)=(0,\pm2)$ .

If $x\ge 1$ , then $1+2^x+2^{2x+1}$ is odd. Hence $8|(y^2-1) = 2^x(2^{x+1} +1)$ , so that $x\ge 3$ .

From $(y-1)(y+1) = 2^x(2^{x+1}+1)$ and the $\gcd(y-1,y+1)=2$ since they are consecutive integers,

we get $\frac{y-1}{2}\times\frac{y+1}{2}= 2^{x-2}(2^{x+2}+1)=k(8k+1)$ , where $k=2^{x-2}$ is a positive power of 2.

Since $\frac{y-1}{2}$ and $\frac{y+1}{2}$ are consecutive integers, there must exist positive odd integers $r,s$ such that $\big\{ \frac{y-1}{2},\frac{y+1}{2}\big\}=\big\{kr,s\big\}$ and $rs=8k+1$ .

Since $1=|kr-s| = |kr -\frac{8k+1}{r}|$ , we must have $r=3$ , Hence $|k-1|=3$ and $2^{x-2}=k=4.$ Thus, $x=4$ and $y=\pm 23.$

The only solutions in integer pairs are $(x,y) \text { are } (0, \pm 2) \text{ and } (4,\pm 23)\implies y_{1}^{x_{1}}+y_{2}^{x_{2}}+ \cdot\cdot\cdot +y_{n}^{x_{n}}= 2^0+(-2)^0+23^4+(-23)^4=\boxed{559684}$ .