We will use the well-known Lifting the Exponent Lemma (LTE), stated as follows:
Lifting the Exponent Lemma: Let $p$ be a prime and $m$ a positive integer. Let $a, b$ be integers such that $p \nmid ab$ and $q_p \mid (a-b)$, where $q_p = p$ when $p$ is odd and $q_2 = 4$ when $p = 2$. Then we have
$$v_p(a^m - b^m) = v_p(a - b) + v_p(m).$$

Assume there exist positive integers $x, y$ satisfying $(2^x - 1)(5^x - 1) = y^n$. We first prove the following lemma:
Lemma: We have $2 \cdot 3 \cdot 5 \cdot 7 \mid x$, and for any prime $p \le n$,
$$v_p(x) \ge \lfloor \log_p n \rfloor.$$

Proof of Lemma: From $2 \mid (5^x - 1)$ we get $2 \mid y$. By LTE:
$$\begin{aligned} n \le nv_2(y) &= v_2((2^x - 1)(5^x - 1)) \\ &= v_2(5 - 1) + v_2(x). \end{aligned}$$
Thus $v_2(x) \ge n - 2 \ge \lfloor \log_2 n \rfloor$. In particular, $4 \mid x$, so $3 \cdot 5 \mid 2^x - 1$, hence $3 \cdot 5 \mid y$. By LTE:
$$\begin{aligned} n \le nv_3(y) &= v_3(((2^2)^{\frac{x}{2}} - 1)((5^2)^{\frac{x}{2}} - 1)) \\ &= v_3(2^2 - 1) + v_3(\frac{x}{2}) + v_3(5^2 - 1) + v_3(\frac{x}{2}) = 2v_3(x) + 2; \\ n \le nv_5(y) &= v_5(((2^4)^{\frac{x}{4}} - 1)(5^x - 1)) \\ &= v_5(2^4 - 1) + v_5(\frac{x}{4}) = v_5(x) + 1. \end{aligned}$$
Thus $v_3(x) \ge (n - 2)/2 \ge \lfloor \log_3 n \rfloor$, and $v_5(x) \ge n - 1 \ge \lfloor \log_5 n \rfloor$. In particular, $3 \mid x$. Combined with $2 \mid x$, we have $6 \mid x$, so $7 \mid (2^x - 1)$, hence $7 \mid y$. By LTE:
$$\begin{aligned} n \le nv_7(y) &= v_7(((2^6)^{\frac{x}{6}} - 1)((5^6)^{\frac{x}{6}} - 1)) \\ &= v_7(2^6 - 1) + v_7(\frac{x}{6}) + v_7(5^6 - 1) + v_7(\frac{x}{6}) = 2v_7(x) + 2. \end{aligned}$$
Thus $v_7(x) \ge (n - 2)/2 \ge \lfloor \log_7 n \rfloor$.

If $n \le 10$, the lemma is already proved. Now assume $n \ge 11$. We proceed by induction on $p$. Let $p$ be a prime with $11 \le p \le n$, and assume the lemma holds for all primes $q < p$. By induction hypothesis, for any prime $q \le p-1$,
$$v_q(x) \ge \lfloor \log_q n \rfloor \ge \lfloor \log_q (p-1) \rfloor \ge v_q(p-1).$$
Thus $(p-1) \mid x$. By Fermat's Little Theorem, $p \mid (2^{p-1}-1)$ and $p \mid (5^{p-1}-1)$. Hence $p \mid (2^x - 1)$ and $p \mid y$. By LTE:
$$\begin{align*} n \le nv_p(y) &= v_p\left(((2^{p-1})^{\frac{x}{p-1}} - 1)((5^{p-1})^{\frac{x}{p-1}} - 1)\right) \\ &= v_p(2^{p-1} - 1) + v_p\left(\frac{x}{p-1}\right) + v_p(5^{p-1} - 1) + v_p\left(\frac{x}{p-1}\right) \\ &= 2v_p(x) + v_p(2^{p-1} - 1) + v_p(5^{p-1} - 1) \\ &< 2v_p(x) + \log_p(2^{p-1}) + \log_p(5^{p-1}) \\ &< 2v_p(x) + p \log_p(10). \end{align*}$$
Therefore,
$$v_p(x) > \frac{n - p \log_p(10)}{2}.$$
Since $11 \le p \le n$, this implies $v_p(x) > 0$, so $v_p(x) \ge 1$. If $n < p^2$, then $v_p(x) \ge 1 \ge \lfloor \log_p n \rfloor$. To complete the induction, we only need to prove for $n \ge p^2 \ge 11^2$ that
$$\frac{n - p \log_p(10)}{2} > \log_p n.$$
This is equivalent to
$$\frac{p^n}{10^p} > n^2.$$
Indeed, using $n \ge p^2 \ge 11^2$, we have
$$\frac{p^n}{10^p} = \frac{p^p}{10^p} \cdot p^{n-p} > 11^{n-\sqrt{n}} > 9^{\frac{n}{2}} > n^2.$$
This completes the induction. The lemma is proved.

Proof of the Main Result: By the lemma, for any prime $p \le n$,
$$v_p(x) \ge \lfloor \log_p n \rfloor \ge v_p(n).$$
Thus $n \mid x$. Let $x = nx_1$, then
$$y^n = (2^x - 1)(5^x - 1) < (10^{x_1})^n.$$
Hence $10^{x_1} \ge y + 1$. Therefore,
$$\begin{align*} y^n &= (10^{x_1})^n - 2^x - 5^x + 1 \\ &\ge (y+1)^n - 2^x - 5^x + 1 \\ &> y^n + n \cdot y^{n-1} - 2 \cdot 5^x \\ &\ge y^n + 4 \cdot y^{\frac{3n}{4}} - 2 \cdot 5^x. \end{align*}$$

This implies $y^{\frac{3n}{4}} < 5^x$, i.e., $y^n < 5^{\frac{4x}{3}}$. On the other hand, by the lemma we have $x \ge 2 \cdot 3 \cdot 5 \cdot 7$, so
$$y^n = (2^x - 1)(5^x - 1) > (0.9 \cdot 2^x) \cdot (0.9 \cdot 5^x) > 0.8 \cdot 5^{\frac{2x}{5}} \cdot 5^x > 5^{\frac{4x}{3}},$$
a contradiction.
Therefore, the original Diophantine equation has no positive integer solutions. $\square$