Note that $$x^2+1=\log_{2}(x+2)-2x$$$$\iff x^2+2x+1=\log_{2}(x+2)$$$$\iff (x+1)^2=\log_{2}(x+2)$$Note that since $(x+1)^2 \ge 0$, we have $x+2\ge 1\implies x\ge -1$. At $x=-1$, both functions intersect at $(-1,0)$. But, it's clear by graphing that $(x+1)^2$ grows faster than $\log_{2}(x+2)$ after $x=-1$, giving no further solutions. $\boxed{x=-1}$