How to prove it has a $\chi^{2}$ distribution I tried to make $T$ close to
$$T_1=\left(\frac{(x_1 - \mu_1)^2}{\sigma_1^2}+\frac{(x_2 - \mu_2)^2}{\sigma_2^2}-2 \frac{\rho}{\sigma_1 \sigma_2} \frac{x_1 - \mu_1}{\sigma_1}\frac{x_2 - \mu_2}{\sigma_2}\right)\frac{1}{1-(\frac{\rho}{\sigma_1\sigma_2})^2}$$
to solve the following problem.
Problem: let $x=(X_1,X_2)'$ follow $N_2(\mu,\Sigma)$, where $\mu=(\mu_1,\mu_2)$, and $\Sigma=\begin{bmatrix}
\sigma_1^2 & \rho\\
\rho & \sigma_2^2
\end{bmatrix}$.
Show that
$$T=\frac{1}{2(1- \rho^2)}\left(\frac{(x_1 - \mu_1)^2}{\sigma_1^2}+\frac{(x_2 - \mu_2)^2}{\sigma_2^2}-2 \rho \frac{x_1 - \mu_1}{\sigma_1}\frac{x_2 - \mu_2}{\sigma_2}\right)$$
has a $\chi^2$ distribution. And give the its parameters?
Since $T_1$ follows as $\chi^2$, I tried to make $T$ close to $T_1$, but I failed since
$$2 \frac{\rho}{\sigma_1 \sigma_2} \frac{x_1 - \mu_1}{\sigma_1}\frac{x_2 - \mu_2}{\sigma_2}\text{ and }2 \rho \frac{x_1 - \mu_1}{\sigma_1}\frac{x_2 - \mu_2}{\sigma_2}$$
are different.

If $X=(X_1,X_2)'$ is normal with mean $\mu=(\mu_1,\mu_2)$, then every quadratic form $$T=a_1(X_1-\mu_1)^2+a_2(X_2-\mu_2)^2+2b(X_1-\mu_1)(X_2-\mu_2),$$ with $a_1\gt0$ and $a_2\gt0$, follows a generalized $\chi^2$ distribution. Only, the parameters depend on $(a_1,a_2,b)$ and on $\Sigma$ the covariance matrix of $X$.
Edit: Assume without loss of generality that $\mu=(0,0)$, consider the matrices $A=\begin{pmatrix}a_1&b\\b&a_2\end{pmatrix}$ and $L$ such that $\Sigma=LL'$, then $X=LY$ where $Y$ is standard normal and $T=X'AX=Y'MY$ where $M=L'AL$ is symmetric. Thus $M=PDP'$ where $P$ is orthogonal and $D$ diagonal, and $T=Z'DZ$ where $Z=P'Y$  is (again) standard normal, that is, $$T=d_1Z_1^2+d_2Z_2^2,$$ where $d_1$ and $d_2$ are the diagonal entries of $D$.
If $d_1$ or $d_2$ is zero, then $T$ is a multiple of a $\chi^2(1)$ random variable. If $d_1$ and $d_2$ are both positive, then a change of variables shows that $T$ is a generalized $\chi^2(2)$ random variable, since the density of $T$ is 
$$f_T(t)=\int_0^{2\pi}\mathrm e^{-((\cos^2 s)/d_1+(\sin^2s)/d_2)t/2}\frac{\mathrm ds}{2\pi\sqrt{d_1d_2}},$$ which can be simplified into
$$f_T(t)=\sqrt{\alpha^2-\beta^2}\,\mathrm e^{-\alpha t}\,I_0\left(\beta t\right),$$ where $I_0$ is the modified Bessel function of the first kind and $$\alpha=\frac14\left(\frac1{d_1}+\frac1{d_2}\right),\qquad\beta=\frac14\left(\frac1{d_1}-\frac1{d_2}\right).$$ In the end:

The key parameters of the problem are the eigenvalues of the matrix $M=L'AL$, where $A$ depends on $(a_1,a_2,b)$, and $\Sigma=LL'$. 

The case in your post reads $$T=\frac{1}{2(1- \rho^2)}\left(\frac{(x_1 - \mu_1)^2}{\sigma_1^2}+\frac{(x_2 - \mu_2)^2}{\sigma_2^2}-2 \rho \frac{x_1 - \mu_1}{\sigma_1}\frac{x_2 - \mu_2}{\sigma_2}\right)$$ and is equivalent to $\Sigma=\begin{pmatrix}1&\rho\\\rho&1\end{pmatrix}$ and $A=\frac{1}{2(1- \rho^2)}\begin{pmatrix}1&-\rho\\-\rho&1\end{pmatrix}$.
(Note that your $\Sigma$, before normalization, should read $\Sigma=\begin{pmatrix}\sigma_1^2&\rho\sigma_1\sigma_2\\\rho\sigma_1\sigma_2&\sigma_2^2\end{pmatrix}$, otherwise $\Sigma$ is not a covariance matrix when $\sigma_1^2\sigma_2^2\lt\rho^2$.)
One can use $L=\begin{pmatrix}1&0\\\rho&\tau\end{pmatrix}$ with $\tau^2=1-\rho^2$, then a simple computation yields $M=\frac12I$ hence $T=\frac12(Y_1^2+Y_2^2)$ where $(Y_1,Y_2)$ is i.i.d. standard normal. Finally, $d_1=d_2=\frac12$ hence $\alpha=1$ and $\beta=0$, thus, the distribution of $T$ is standard exponential, with $$f_T(t)=\mathrm e^{-t}.$$
