This question is based on Every natural number is covered by consecutive numbers that sum to a prime power.
Let $T(n) = \frac{n(n+1)}2$ be the $n$th triangular number, and let $p^j$ denote a prime power.
By checking $1 \leq n \leq 10\,000$, it appears empirically that after fixing $k$, $T(n)-T(k)$ is a prime power only for a finite number of values $n$:$$ \begin{align*} T(n) - T(0) = p^j &\Longrightarrow n \in \{ 2 \}\\ T(n) - T(1) = p^j &\Longrightarrow n \in \{ 2,3,4,7 \}\\ T(n) - T(2) = p^j &\Longrightarrow n \in \{ 3,4,7 \}\\ T(n) - T(3) = p^j &\Longrightarrow n \in \{ 4,5,10 \}\\ T(n) - T(4) = p^j &\Longrightarrow n \in \{ 5,6,13,22 \}\\ T(n) - T(5) = p^j &\Longrightarrow n \in \{ 7,16 \}\\ T(n) - T(6) = p^j &\Longrightarrow n \in \{ 7,19 \}\\ T(n) - T(7) = p^j &\Longrightarrow n \in \{ 8,9,10,17 \}\\ T(n) - T(8) = p^j &\Longrightarrow n \in \{ 9,10,25 \}\\ T(n) - T(9) = p^j &\Longrightarrow n \in \{ 28 \}. \end{align*} $$
Is it easy to prove that each of these sets is finite? If so, is there a way to compute an upper bound for the largest number that can appear in one of the sets, or otherwise compute the size of each set?
Yes, it's true these sets will be finite for any $k$, and the proof below shows how to compute a reasonable upper bound on the largest number which can appear. To see this, as I showed in my answer to the question you linked to as yours' being based on, you have
$$T(n) - T(k) = \frac{(n + k + 1)(n - k)}{2} = p^j \tag{1}\label{eq1A}$$
Apart from a factor of $2$ in either $n + k + 1$ or $n - k$, the only prime factor of these $2$ expressions is $p$ for some prime $p$. Assume $n - k \gt 2$, so it has at least one factor of $p$, to get
$$n - k \equiv 0 \pmod p \implies n \equiv k \pmod p \tag{2}\label{eq2A}$$
Thus, you also have
$$\begin{equation}\begin{aligned} n + k + 1 & \equiv 0 \pmod p \\ k + k + 1 & \equiv 0 \pmod p \\ 2k + 1 & \equiv 0 \pmod p \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows $p$ is limited to the prime factors of $2k + 1$.
Consider that $2 \mid n + k + 1$. Assuming $n - k \gt 1$, you have for some positive integers $q$ and $r$ (where $q + r = j$, so it works in \eqref{eq1A}) that
$$n + k + 1 = 2p^q \tag{4}\label{eq4A}$$
$$n - k = p^r \tag{5}\label{eq5A}$$
Next, \eqref{eq4A} minus \eqref{eq5A} gives
$$2k + 1 = 2p^q - p^r \tag{6}\label{eq6A}$$
If $s = \min(q,r)$, then $p^s \mid 2k + 1$. However, since $2k + 1$ is a fixed value, there's a maximum value allowed for $s$. This limits the maximum possible value for $n$ in \eqref{eq4A} and \eqref{eq5A}. You can also do basically the same analysis in the case where $2 \mid n - k$ instead.
In summary, this shows since for any prime factor $p$ of $2k + 1$ there's a maximum possible $n$ which works, and there are a finite number of these prime factors $p$, there are thus at most a finite number of solutions to \eqref{eq1A} for any particular $k$.