This is Exercise 1.4 in Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory.
Write $\alpha = \sqrt[3]{22}$ and $k = \mathbb{Q}(\alpha)$. I have shown that $k$ has class number 3 and the norm form is $$ \mathrm{norm}(a + b\alpha + c\alpha^2) = a^3 + 22b^3 + 22^2c^3 - 66abc. $$
Assuming that $(x,y,z)$ is a primitive solution of $X^3+22Y^3+3Z^3=0$ with $z > 0$, we find $$ N(x+y\alpha) = |x^3+22y^3| = 3z^3. $$ Since $3$ is totally ramified in $k$, i.e. $3\mathcal{O}_k = \mathfrak{p}_3^3$, we can write $$ (x+y\alpha) = \mathfrak{p}_3 \mathfrak{a}, $$ for some $\mathfrak{a}$ with $N\mathfrak{a} = z^3$. The exercise says to show that $\mathfrak{a}$ is a cube, which would lead to a contradiciton since $\mathfrak{p}_3$ is not principal, and this is where I am having trouble.
Writing $\mathfrak{a} = \prod \mathfrak{p}^{a_{\mathfrak{p}}}$, the primes with $a_{\mathfrak{p}} \neq 0$ are above the primes dividing $z$ and I want to show that $a_{\mathfrak{p}} \equiv 0 \bmod 3$ for every $\mathfrak{p}$. If $N\mathfrak{p} = p^{f_{\mathfrak{p}}}$, the fact that the norm of $\mathfrak{a}$ is a cube says $$ \sum_{\mathfrak{p} \mid p} a_{\mathfrak{p}} f_{\mathfrak{p}} \equiv 0 \bmod 3.$$
If $p = 2,3$ or $11$, then $p$ is totally ramified in $k$, i.e. $p\mathcal{O}_k = \mathfrak{p}_p^3$, so $p \mid z$ implies $\mathfrak{p}_p^3 \mid \mathfrak{a}$ and the result holds.
I now consider a prime $p \neq 2,3,11$ which divides $z$. The equation gives $$ x^3 + 22y^3 \equiv 0 \bmod p^3, $$ and because of coprimality, we find that $22$ is a cube modulo $p$ so we cannot have $f_{\mathfrak{p}} = 3$ for any $\mathfrak{p} \mid p$. This gives $a_{\mathfrak{p}} \equiv 0 \bmod 3$ when $\mathfrak{p}$ is the only prime above $p$ dividing $\mathfrak{a}$.
The problem is that I do not see how to reach that conclusion when there are more than one prime factor of $p\mathcal{O}_k$ appearing in the decomposition of $\mathfrak{a}$. For instance, if $p\mathcal{O}_k = \mathfrak{p}_1 \mathfrak{p}_2 \mathfrak{p}_3$, why can we not have $\mathfrak{a} = \mathfrak{p}_1 \mathfrak{p}_2^2$?
See here for a treatment of the similar (and more famous) equation $3x^3 + 4y^3 + 5z^3 = 0$. This example and the one you write about are also in Chapter 10 of Cassels' Local Fields: see Lemma 9.1 and 9.2.