Given a field $K$, let $\alpha_1,\alpha_2,\cdots,\alpha_n \in K^*$ and $a_1,\cdots ,a_n \in K $. Show that if $a_1\alpha_1^m+a_2\alpha_2^m+\cdots+a_n\alpha_n^m =0$ for all $m \in \Bbb Z$, then $a_1 = \cdots = a_n = 0.$
From here I have considered a system of $n$ equations putting $m = 0,1,...,n-1$ and considered the matrix and claimed that the det is non zero.
Your approach is essentially correct. The linear system of equations
$$\left\{ \begin{eqnarray} a_1 &+& a_2 &+& \dots &+& a_n &=& 0\\ \alpha_1 a_1 &+& \alpha_2 a_2 &+& \dots &+& \alpha_n a_n &=& 0\\ \alpha_1 ^2 a_1 &+& \alpha_2 ^2 a_2 &+& \dots &+& \alpha_n ^2 a_n &=& 0\\ \vdots\\ \alpha_1 ^{n-1} a_1 &+& \alpha_2 ^{n-1} a_2 &+& \dots &+& \alpha_n ^{n-1} a_n &=& 0\\ \end{eqnarray} \right.$$
having the unknowns $a_1, \dots, a_n$ has for determinant the Vandermonde determinant
$$\begin{vmatrix} 1 & 1 & \dots & 1 \\ \alpha_1 & \alpha_2 & \dots & \alpha_n \\ \alpha_1 ^2 & \alpha_2 ^2 & \dots & \alpha_n ^2 \\ \vdots\\ \alpha_1 ^{n-1} & \alpha_2 ^{n-1} & \dots & \alpha_n ^{n-1} \\ \end{vmatrix} = \prod _{1 \le i, j \le n} (\alpha_i - \alpha_j) .$$
If $\alpha_i \ne \alpha_j$ for $i \ne j$, then the determinant is non-zero, hence the solution $a_i = 0$ for all $1 \le i \le n$ is the only one.
If there exist $i$ and $j$ with $\alpha_i = \alpha_j$, then the determinant is $0$, hence there exist non-trivial solutions.
Notice that allowing $m$ to be arbitrary in $\Bbb Z$ was way too much, since we only used $m \in \{0, \dots, n-1\}$.
(Assuming the $\alpha_i$'s are distinct:)
The given system of equations implies that $$a_1 p(\alpha_1) + a_2 p(\alpha_2) + \cdots + a_n p(\alpha_n) = 0$$ for any polynomial $p$ with coefficients in $K$. Choosing, for example, $$p(x) = (x-\alpha_2)(x-\alpha_3)\cdots (x-\alpha_n),$$ we see that $a_1 p(\alpha_1) = 0 \implies a_1 = 0$, since the $\alpha_i$'s are distinct. Similarly, we see $a_2 = a_3 = \cdots = a_n = 0$.