We first show a solution for each $n \in \{2,3,4\}$. We will later show the impossibility of finding such a solution for $n \geqslant 5$. For $n=2$, take for example $(a_{1}, a_{2}) = (1,3)$ and $r=2$. For $n=3$, take the root $r > 1$ of $x^{2} - x - 1 = 0$ (the golden ratio) and set $(a_{1}, a_{2}, a_{3}) = (0, r, r + r^{2})$. Then
$$(a_{2} - a_{1}, a_{3} - a_{2}, a_{3} - a_{1}) = (r, r^{2}, r + r^{2} = r^{3})$$
For $n=4$, take the root $r \in (1,2)$ of $x^{3} - x - 1 = 0$ (such a root exists because $1^{3} - 1 - 1 < 0$ and $2^{3} - 2 - 1 > 0$) and set $(a_{1}, a_{2}, a_{3}, a_{4}) = (0, r, r + r^{2}, r + r^{2} + r^{3})$. Then
$$(a_{2} - a_{1}, a_{3} - a_{2}, a_{4} - a_{3}, a_{3} - a_{1}, a_{4} - a_{2}, a_{4} - a_{1}) = (r, r^{2}, r^{3}, r^{4}, r^{5}, r^{6})$$
For $n \geqslant 5$, we will proceed by contradiction. Suppose there exist numbers $a_{1} < a_{2} < \cdots < a_{n}$ and $r > 1$ satisfying the conditions of the problem. We start with a lemma:

Lemma. We have $r^{n-1} > 2$.

Proof. There are only $n-1$ differences $a_{j} - a_{i}$ with $j = i+1$, so there exists an exponent $e \leqslant n$ and a difference $a_{j} - a_{i}$ with $j \geqslant i+2$ such that $a_{j} - a_{i} = r^{e}$. This implies
$$r^{n} \geqslant r^{e} = a_{j} - a_{i} = (a_{j} - a_{j-1}) + (a_{j-1} - a_{i}) > r + r = 2r,$$
thus $r^{n-1} > 2$ as desired.

To illustrate the general approach, we first briefly sketch the idea behind the argument in the special case $n=5$. In this case, we clearly have $a_{5} - a_{1} = r^{10}$. Note that there are 3 ways to rewrite $a_{5} - a_{1}$ as a sum of two differences, namely
$$(a_{5} - a_{4}) + (a_{4} - a_{1}), (a_{5} - a_{3}) + (a_{3} - a_{1}), (a_{5} - a_{2}) + (a_{2} - a_{1}).$$
Using the lemma above and convexity of the function $f(n) = r^{n}$, we argue that those three ways must be $r^{10} = r^{9} + r^{1} = r^{8} + r^{4} = r^{7} + r^{6}$. That is, the "large" exponents keep dropping by 1, while the "small" exponents keep increasing by $n-2, n-3, \ldots, 2$. Comparing any two such equations, we then get a contradiction unless $n \leqslant 4$.

Now we go back to the full proof for any $n \geqslant 5$. Denote $b = \frac{1}{2} n(n-1)$. Clearly, we have $a_{n} - a_{1} = r^{b}$. Consider the $n-2$ equations of the form:
$$a_{n} - a_{1} = (a_{n} - a_{i}) + (a_{i} - a_{1}) \text{ for } i \in \{2, \ldots, n-1\}$$
In each equation, one of the two terms on the right-hand side must be at least $\frac{1}{2} (a_{n} - a_{1})$. But from the lemma we have $r^{b-(n-1)} = r^{b} / r^{n-1} < \frac{1}{2} r^{b}$, so the other term must be at least $\frac{1}{2} (a_{n} - a_{1})$. Therefore, we can write
$$a_{n} - a_{1} = r^{b} = r^{\alpha_{i}} + r^{\beta_{i}}$$
for some $\alpha_{i}$ and $\beta_{i}$ with $\alpha_{i} \geqslant b - (n-1)$ and $\beta_{i} \leqslant b - (n-1)$. Since $r > 1$ and $f(r) = r^{n}$ is convex, we have
$$r^{b-1} - r^{b-2} > r^{b-2} - r^{b-3} > \ldots > r^{b-(n-3)} - r^{b-(n-2)}$$
implying
$$r^{\alpha_{2}} - r^{\alpha_{1}} > r^{\alpha_{3}} - r^{\alpha_{2}} > \ldots > r^{\alpha_{n-2}} - r^{\alpha_{n-3}}.$$
Convexity of $f(r) = r^{n}$ further implies
$$\alpha_{2} - \alpha_{1} > \alpha_{3} - \alpha_{2} > \ldots > \alpha_{n-2} - \alpha_{n-3}.$$
Note that $\alpha_{n-2} - \alpha_{n-3} \geqslant 2$: Otherwise we would have $\alpha_{n-2} - \alpha_{n-3} = 1$ and thus
$$r^{\alpha_{n-3}} \cdot (r-1) = r^{\alpha_{n-2}} - r^{\alpha_{n-3}} = r^{b-(n-3)} - r^{b-(n-2)} = r^{b-(n-2)} \cdot (r-1),$$
implying that $\alpha_{n-3} = b - (n-2)$, a contradiction. Therefore, we have
$$\begin{aligned} \alpha_{n-2} - \alpha_{1} & = (\alpha_{n-2} - \alpha_{n-3}) + \cdots + (\alpha_{2} - \alpha_{1}) \\ & \geqslant 2 + 3 + \cdots + (n-2) \\ & = \frac{1}{2} (n-2)(n-1) - 1 = \frac{1}{2} n(n-3) \end{aligned}$$
On the other hand, from $\alpha_{n-2} \leqslant b - (n-1)$ and $\alpha_{1} \geqslant 1$ we get
$$\alpha_{n-2} - \alpha_{1} \leqslant b - n = \frac{1}{2} n(n-1) - n = \frac{1}{2} n(n-3),$$
implying that equalities must occur everywhere and the claim about the small terms follows. Now, assuming $n-2 \geqslant 2$, we have the two different equations:
$$r^{b} = r^{b-(n-2)} + r^{b-(n-2)-1} \text{ and } r^{b} = r^{b-(n-3)} + r^{b-(n-2)-3},$$
which can be rewritten as
$$r^{n-1} = r + 1 \quad \text{ and } \quad r^{n+1} = r^{4} + 1$$
Simple algebra now gives
$$r^{4} + 1 = r^{n+1} = r^{n-1} \cdot r^{2} = r^{3} + r^{2} \Longrightarrow (r-1)(r^{3} - r - 1) = 0.$$
Since $r \neq 1$, using Equation (1) we conclude $r^{3} = r + 1 = r^{n-1}$, thus $n = 4$, which gives a contradiction.