The equation $$ \frac{x^k-1}{x-1}=y^q$$ is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $q\geq 2$, the only solutions are $$ \frac{3^5-1}{3-1}=11^2,\qquad \frac{7^4-1}{7-1}=20^2,\qquad \frac{18^3-1}{18-1}=7^3.$$ For $q=2$, the equation was solved by Ljunggren (Norsk. Mat. Tidsskr. 25 (1943), 17-20).

Hence for your problem (where $q=2$ and $x$ is a prime power) we can assume that $k=1$ or $k=2$. In the first case, $y=1$ and $x$ is arbitrary, which we can regard as trivial solutions. In the second case, $x=y^2-1$ is a prime power, whence it is easy to see that either $x=8$ and $y=3$, or $x=3$ and $y=2$. To summarize, your equation only has the four solutions listed in your post, besides the trivial solutions ($k=1$ and $y=1$).

The equation $$ \frac{x^k-1}{x-1}=y^m$$ is known as the Nagell-Ljunggren equation. It is conjectured that for $x\geq 2$, $y\geq 2$, $k\geq 3$, $m\geq 2$, the only solutions are $$ \frac{3^5-1}{3-1}=11^2,\qquad \frac{7^4-1}{7-1}=20^2,\qquad \frac{18^3-1}{18-1}=7^3.$$ For $m=2$, the equation was solved by Ljunggren (Norsk. Mat. Tidsskr. 25 (1943), 17-20).

Hence for your problem (where $m=2$ and $x$ is a prime power) we can assume that $k=1$ or $k=2$. In the first case, $y=1$ and $x$ is arbitrary, which we can regard as trivial solutions. In the second case, $x=y^2-1$ is a prime power, whence it is easy to see that either $x=8$ and $y=3$, or $x=3$ and $y=2$. To summarize, your equation only has the four solutions listed in your post, besides the trivial solutions ($k=1$ and $y=1$).