Let $t=2013$, then the original question is equivalent to showing $t^2 + t^2\times (t+1)^2+(t+1)^2$ is a perfect square.

\begin{align*}&t^2 + t^2\times (t+1)^2+(t+1)^2\\&=t^2 + t^2(t^2 + 2t+1) + (t^2+2t+1)\\&=t^4 + 2t^3+3t^2+2t+1\\&=t^2\left(t^2+2t+3+\frac{2}{t} + \frac{1}{t^2}\right)\\&=t^2\left(\left(t^2+\frac{1}{t^2}+2\right) + 2\left(t+\frac{1}{t}\right) + 1\right)\\&=t^2\left(\left(t+\frac{1}{t}\right)^2+2\left(t+\frac{1}{t}\right)+1\right)\\&=t^2\left(t+\frac{1}{t}+1\right)^2\\&=(t^2+1+t)^2\end{align*}