The fundamental solution of the Pell equation $$x^{2}-3y^{2}=1$$ is $2+\sqrt{3}$.
It seems to me that if $x_{n}+y_{n}\sqrt{3}$, $x_{n}, y_{n} \in \mathbb{N}$, is a solution of the said Pell equation and $x_{n}$ is a power of $7$, then $n=2$.
Simple congruence arguments allow us to conclude that such an $n$ can't be an odd natural number or divisible by $4$.
Do you see a way to show that $n$ cannot be congruent to $2$ modulo $4$ unless $n=2$?
We know $x_n + y_n\sqrt{3} = (2+\sqrt{3})^n$. If $n=4k+2$ for some $k\ge 0$, then $$x_n + y_n\sqrt{3} = (2+\sqrt{3})^n = (7+4\sqrt{3})^{2k+1}, $$ which implies that $$x_n = \sum_{i=0}^{k}C(2k+1, 2i) 7^{2k+1-2i}(4\sqrt{3})^{2i} = \sum_{i=0}^{k}C(2k+1, 2i) 7^{2k+1-2i}48^i$$
We need to consider how many factors of $7$ each $C(2k+1, 2i)$ contains. I don't have time now and will claim that: if $7^\alpha || 2k+1$, we have $7^{\alpha + 1} \mid 7^{2k-2i}C(2k+1, 2i)$ for $0\le i<k$. The proof uses the factors of a prime in a factorial expressed in the sum of floor functions.
If the claim holds, then $7^{\alpha + 2}$ divides all terms but the last one.
Claim: $m>n>0$, $p$ is a prime number, $\alpha >0$ such that $p^\alpha || m$ (that is, $p^\alpha \mid m$ and $p^{\alpha+1} \nmid m$). If $p^\beta || C(m, n)$, then $\beta \ge \alpha - s$, where $s\ge 0$ and $p^s \le n$ and $p^{s+1} < n$.
Proof of claim: $$\beta =\sum_{i=1}^{\infty} ([\frac{m}{p^i}] - [\frac{n}{p^i}] - [\frac{m-n}{p^i}])$$
Each term in the summation contributes $0$ or $1$. If $n=1$, the first $\alpha$ terms contribute $1$ each; for general $n$, except the first $s$ terms, the contribution of each term is at least as good as in $C(m, 1)$.