Let $9x^2 - 40x + 39 = p^n$ for some prime $p$ and some non-negative integer $n$. From
$$p^n = 9x^2 - 40x + 39 = (9x - 13)(x - 3)$$
it follows that $9x - 13 = p^k$ and $x - 3 = p^l$ or $9x - 13 = -p^k$ and $x - 3 = -p^l$ for some integers $k$ and $l$, where $0 \le l < k$ and $n = k + l$.

First, let us solve the system of equations $9x - 13 = p^k$ and $x - 3 = p^l$. We have $9(p^l + 3) - 13 = p^k$ or $14 = p^k - 9p^l = p^l(p^{k-l} - 9)$. If $l = 0$, then $p^k = 23$, so $p = 23$, $k = 1$ and $x = 4$. Else, we have $l \ge 1$ and $p^l = 14$, so either $p^l = 2$ and $p^{k-l} - 9 = 7$ or $p^l = 7$ and $p^{k-l} - 9 = 2$. In the first case we have $p = 2$, $l = 1$, $k = 5$ and $x = 5$. In the second case there are no solutions.

Now, consider the other system of equations, $9x - 13 = -p^k$ and $x - 3 = -p^l$. Here we have $14 = p^l(9 - p^{k-l})$. The only two possibilities are $p = 2$, $l = 1$ and $p = 7$, $l = 1$ (in this case we cannot have $l = 0$). We get $x = 1$ and $x = -4$.