HMMT Invitational Competition

Solution:
In general we can assume that each $a_{i}>1$, since replacing $a_{i}=1$ by some large integer $a$ creates a set $T$ containing the original $T$ as a subset (by setting $e_{i}=0$).

We proceed by induction on $n$. For the base case $n=1$, an arithmetic progression of length at least $3$ would give $a_{1}^{e_{1}}+a_{1}^{e_{3}}=2 a_{1}^{e_{2}}$ where $e_{3}>e_{2}>e_{1}$. But $a_{1}^{e_{3}} \geq 2 a_{1}^{e_{2}}$, so this is impossible. Thus our result holds for $n=1$.

Assume the result is true for some $n-1$, and let $A_{n-1}$ be a number such that the longest progression when $|S|=n-1$ has length less than $A_{n-1}$. Let $M$ be a large integer that we will choose later. Take an arithmetic progression of length $2M+1$, calling the terms $b_{1}, b_{2}, \ldots, b_{2M+1}$. Note that $b_{2M+1} \leq 2 b_{M+1}$, since $b$ is a sequence of positive integers. For each term from $b_{M+1}$ to $b_{2M+1}$ assign to it the maximum power that is part of the sum. Call this value $c(b_{i})$. More explicitly, if $b_{i}=\sum_{j=1}^{n} a_{j}^{e_{j}}$, then $c(b_{i})=\max \left(a_{1}^{e_{1}}, a_{2}^{e_{2}}, \ldots, a_{n}^{e_{n}}\right)$.

Since $\frac{b_{M+1}}{n} \leq c(b_{i}) \leq b_{2M+1} \leq 2 b_{M+1}$ for $M+1 \leq i \leq 2M+1$, there are at most $\sum_{j=1}^{n}\left(\log_{a_{i}} 2n+1\right) \leq n\left(\log_{2} 2n+1\right)$ different values of $c(b_{i})$. By Van der Waerden's Theorem, there exists a value of $M$ such that coloring an arithmetic progression of length $M$ with $n\left(\log_{2} 2n+1\right)$ colors yields a monochromatic arithmetic progression of length $A_{n-1}$. In particular, we can take $M=W\left(A_{n-1}, n\left(\log_{2} 2n+1\right)\right)$, where $W(n, k)$ denotes the Van der Waerden number. So, we color $b_{M+1}, \ldots, b_{2M+1}$ by their $c(b_{i})$. Subtracting the common perfect power from each term of the monochromatic arithmetic progression obtained gives an arithmetic progression of $A_{n-1}$ integers expressible as the sum of perfect powers of distinct numbers in $S \setminus a_{j}$. By the inductive hypothesis, this is a contradiction. So no arithmetic progression of length $2M+1$ can be contained in $T$, and we can take $A_{n}=2M+1$.

By induction, we thus have such an $A_{n}$ for all $n$.