Proving that the Diophantine equation $(11a + 5b)^2 - 223b^2 = \pm 11$ has no solutions

There are techniques due to Dirichlet that achieve what you want in a finite number of steps. In the present case, the following ad-hoc computations do the trick.

First observe that $\alpha = 14 + \sqrt{223}$ has norm $-27$ (this implies that your second equation has a rational solution, which in turn suggests that you cannot prove it impossible by working modulo integers). Thus if there is an element of nurm $\pm 3$, one of the elements $\alpha$, $\varepsilon \alpha$ or $\varepsilon^2\alpha$ must be a cube, where $\varepsilon = 224 + 15 \sqrt{223}$ is the fundamental unit (which can be computed from the element $\beta = 15 + \sqrt{223}$ with norm $2$ via $\varepsilon = \beta^2/2$). Now you check that none of these elements is a cube.

For showing that $\alpha$ is not a cube assume that $\alpha = \gamma^3$ and $\alpha' = {\gamma'}^3$. Then $\gamma \approx 3.07$ and $\gamma' \approx -0,977$, and since $\gamma + \gamma'$ is not an integer, this is impossible.

The ideals of norm $11$ do not contribute to the class group since $16 \pm \sqrt{223}$ have norm $33$.