The system of equations: $\begin{cases} 2x^2=4y^2+3z^2+2; \\ 13x=4y+3z+29 \end{cases}$ Solve in positive integers the system of equations:
$$\begin{cases}
2x^2=4y^2+3z^2+2;
\\
13x=4y+3z+29
\end{cases}$$
My work so far:
I used wolframalpha: $x=3,y=1,z=2$. 

$$2x^2=4y^2+3z^2+2\tag1$$
$$13x=4y+3z+29\tag2$$
(1)
$$2x^2=4y^2+3z^2+2 \Rightarrow x>\max\{y,z\}$$
(2)
$$13x=4y+3z+29<4x+3x+29 \Rightarrow 6x<29 \Rightarrow x<5$$
(1) $z -$ even. (2) $x -$ odd. Hence $x \in \{1,3\}$. But $x>y,z\ge1$. Hence $$x=3$$
