Minimizing a quadratic function subject to quadratic constraints
Hint: Your conclusion that your three equations imply that there is no consistent choice for $\lambda$ only holds if $x$, $y$ and $z$ are non-zero! If any of them are zero, then their corresponding equation tells us nothing about $\lambda$ since it is trivially fulfilled.
You thus know that the critical points are of the form $(x,0,0)$, $(0,y,0)$ and $(0,0,z)$. Up to you to check them out
When you solve the system with Lagrange method, your variables are $x,y,z$ and $\lambda$.
One solution to the system is $$(x,y,z,\lambda)=(1,0,0,\frac{1}{4}),$$ which matches your intuitive solution.
Alternatively, since $z^2=4-4x^2-2y^2$ cannot be negative (thank you @user35734!), it is equivalent to minimize $$f(x,y,z(x,y))=4-3x^2-y^2$$ subject to: $$ 4x^2+2y^2 \le 4 $$
This is an elliptic domain, and it is easy to see that the minimum will be reached on its border, therefore we can replace variables $x$ and $y$ by $\cos t$ and ${\sqrt{2}} \sin t$, respectively. The problem then boils down to minimize $$ f(x(t),y(t))=4-3\cos^2t - 2\sin^2 t $$ Standard calculus techniques yield $$ \min f = 1 $$
To complement the other answers, we are looking for the points on a given ellipsoid that are closest to the origin. Here is the plot a sphere of radius $1$ touching the given ellipsoid at $(\pm 1, 0, 0)$.