You know that $$m\vec a=q\vec v \times \vec B,$$

So in particular, since $\vec B = B\hat z,$ we have $$ma_x=qv_yB,\text{ and } ma_y=-qv_xB. $$

And in our case $B=-\beta x$ so we have $$m\ddot x=-q\dot y\beta x\text{ and } m\ddot y=q\dot x\beta x. $$

You can take the time derivative of the left equation and get $$m\dddot x=-q\ddot y\beta x-q\dot y\beta \dot x,$$ which becomes

$$m\dddot x=-q\left(\frac{q\dot x\beta x}{m}\right)\beta x-q\left(\frac{m\ddot x}{-q\beta x}\right)\beta \dot x.$$ 

This third order equation needs initial conditions $x(0)=0,$ $\dot x(0)=v,$ and $\ddot x(0)=0$ (why?). And you can then find the time $t=T$ when $\dot x(t=T)=0.$ From that $T$ and the function $x(t)$ you can get $x(T)$ which is half of what you want.

You can make a similar third order equation for $y(t)$ and either solve it and then get $t=T$ where $\dot y(t=T)=\pm v$ (which one?) and confirm the two $T$ are the same ($\dot x^2+\dot y^2$, unlike the canonical momentum, is conserved). And in the end from the $T$ and the function $y(t)$ you can get $y(T)$ which is  the other half of what you want.

Note I didn't use a vector potential, a Lagrangian, a Hamiltonian, or an appeal to QM. And the one use of a conserved quantity, $\dot x^2+\dot y^2,$ was optional.

So what's wrong with what you did? Firstly, canonical momentum is usually not a conserved quantity, it is just the variable needed to make a Legendre transformation from the $Q, \dot Q$ coordinates of the Lagrangian to the $Q,P$ coordinates of the Hamiltonian. If the Lagrangian doesn't depend on a $Q$ then the corresponding $P$ is conserved.  Now, $L(\vec x, \vec v)=-mc^2\sqrt{1-\frac{v^2}{c^2}}+q\vec A\cdot\vec v -q\Phi,$ since 

$$\frac {d}{dt}\left(\frac{mv_i}{\sqrt{1-\frac{v^2}{c^2}}}+qA_i\right)=\frac {d}{dt}\left(\frac{\partial L}{\partial v_i}\right)=\frac {d P_i}{dt}$$

and

$$\frac{\partial L}{\partial Q_i}=v_x\frac{\partial A_x}{\partial Q_i}+v_y\frac{\partial A_y}{\partial Q_i}+v_z\frac{\partial A_z}{\partial Q_i}.$$

The right hand side is not zero, so the Euler-Lagrange equations $\frac {d}{dt}\left(\frac{\partial L}{\partial v_i}\right)=\frac{\partial L}{\partial Q_i}$ do not yield that the momentum $P_i=\frac{\partial L}{\partial v_i}$ is conserved and in fact 

$$\frac {d P_i}{dt}=\frac{\partial L}{\partial Q_i}=v_x\frac{\partial A_x}{\partial Q_i}+v_y\frac{\partial A_y}{\partial Q_i}+v_z\frac{\partial A_z}{\partial Q_i}.$$

The reason we know that the Lagrangian is $L(\vec x, \vec v)=-mc^2\sqrt{1-\frac{v^2}{c^2}}+q\vec A\cdot\vec v -q\Phi,$ is because the Euler-Lagrange equations $\frac {d}{dt}\left(\frac{\partial L}{\partial v_i}\right)=\frac{\partial L}{\partial Q_i}$ leads to

$$\frac {d}{dt}\left(\frac{mv_i}{\sqrt{1-\frac{v^2}{c^2}}}+qA_i\right)=v_x\frac{\partial A_x}{\partial Q_i}+v_y\frac{\partial A_y}{\partial Q_i}+v_z\frac{\partial A_z}{\partial Q_i}.$$

Which is equivalent to $$\frac {d}{dt}\left(\frac{m\vec v}{\sqrt{1-\frac{v^2}{c^2}}}\right)=q\left(-\frac{\partial \vec A}{\partial t}-\vec \nabla \Phi + \vec v \times \left(\vec \nabla \times \vec A\right)\right).$$

Since other people brought up Hamiltonians, you can get a Hamiltonian as $H(\vec Q,\vec P,t)=\sqrt{(mc^2)^2+(\vec P-q\vec A)^2c^2}+q\Phi$ (which is approximately $mc^2 +(\vec P-q\vec A)^2/2m +q\Phi$ in the nonrelativistic limit). And we can denote $\vec p=\vec P -q\vec A$ as the kinetic momentum. And for people that bring up quantum theory into this, the canonical momentum is what obeys the canonical commutator relationship so since the kinetic momentum is $\vec p=\vec P -q\vec A$ the different components of the kinetic momentum do not commute with each other because of the vector potential term.