It seems to me that your problem can be much generalized, using more powerful tools from ANT such as Eisenstein's cubic recipocity law. Consider the diophantine equation (1) $x^3+qy^3=pz^3$, where $p,q$ are two distinct given primes, et let us solve it in two steps:

1) Suppose that we can show that (1) implies that $x\equiv y\equiv 0$ mod $p$. Then we can do infinite descent exactly as in the answer of @greedoid to conclude that (1) has no non trivial solution.

2) To go on, let us assume $y \neq 0$ mod $p$. Then the reduction of (1) modulo $p$ is equivalent to $q\equiv t^3$ mod $p$, $t\in \mathbf Z$, and we must distinguish three cases: (i) If $p=3$, Fermat's little theorem says that our congruence always has a solution ; (ii) If $p\equiv -1$ mod $3$, taking cubic powers induces an automorphism of $(\mathbf Z/p\mathbf Z)^*$, so our congruence is again solvable ; (iii) If $p\equiv 1$ mod $3$, the arithmetic of the Eisenstein ring $R=\mathbf Z[\omega]$, where $\omega$ is a primitive 3-rd root of unity, enters the game. It is classically known that $R$ is a PID (and even an euclidian ring), and that $p$ splits as a product $p=\pi \bar\pi$ , where $\pi$ is a prime of $R$. Then our congruence is solvable iff $(\frac q\pi)_3=1$, where $(\frac ..)_3$ denotes the cubic Legendre symbol. For the definition and properties of this symbol, a convenient reference is David Cox's book " Primes of the form $x^2+ny^2$ ", chap. 1, ยง4. *Supposing further that* $q\equiv \pm 1$ mod $3$, the cubic reciprocity law asserts that $(\frac q\pi)_3(\frac \pi q )_3=1$, so that our congruence is solvable iff $(\frac \pi q )_3=1$. By formula (4.10) in *loc. cit.*, this is equivalent to $(\frac \pi q )_3 \equiv \pi ^{(q^2-1)/3}$mod $q$, and finally our congruence (coming from $y\neq 0$ mod $p$) is solvable iff $p\equiv 1$ mod $3$ and $\pi ^{(q^2-1)/3} \equiv 1$ mod $q$.

Summarizing, the infinite descent in 1) can be performed iff $p\equiv 1$ mod $3$ and $\pi ^{(q^2-1)/3} \neq 1$ mod $q$. You can easily check that this criterion contains your special case $p=7, q=2$.

NB: The extra hypothesis $q\equiv \pm 1$ mod $3$ is not necessary, but without it the cubic reciprocity law becomes more complicated to use.