The proof is essentially a size argument. We split into three cases, the first two of which are quite straightforward.

Case 1: two of $p, q, r$ are equal, wlog $q = r$. Subtract $a^q + b^q + c^q$ from the given equation to show that
$$a^p - a^q = b^p - b^q = c^p - c^q.$$
Now if $p > q$ then $x^p - x^q = x^q(x^{p-q} - 1)$ is a strictly increasing function of $x > 0$, so the only way the equality can hold is if $a = b = c$. If $p < q$ then the same argument shows that $a = b = c$ also. Yet the only remaining subcase is when $p = q = r$.

Case 2: two of $a, b, c$ are equal, wlog $b = c$. Subtract $b^p + b^q + b^r$ from the given equation to show that $a^p - b^p = a^q - b^q = a^r - b^r$. Exactly as in the previous case, the function of a positive integer $s$ $a^s - b^s = (a-b)(a^{s-1} + a^{s-2}b + \dots + b^{s-1})$ is strictly increasing, strictly decreasing or zero according as $a > b$, $a < b$ or $a = b$. In the first two subcases, this forces $p = q = r$, and in the last subcase $a = b = c$.

Case 3: the $a, b, c$ are distinct, as are the $p, q, r$. Wlog $a$ is the greatest of $a, b, c$ and (cycling the variables if necessary) $p$ is the greatest of $p, q, r$. In particular, $a, p \ge 3$. We claim that for such $a, p$ we have
$$a^p \ge (a-1)^p + 2a^{p-1}.$$
Indeed, since $(a-1)^p + 2a^{p-1} \le a^{p-3}((a-1)^3 + 2a^2)$ it suffices to prove the inequality for $p=3$, when it rearranges to the inequality $a^2 - 3a + 1 \ge 0$. This certainly holds for $a \ge 3$. As a consequence, we have the inequality
$$a^p + b^q + c^r > (a-1)^p + 2a^{p-1} \ge b^p + c^q + a^r$$
which is a contradiction.