Prove that for every positive integer $m$ there exists a positive integer $n_m$ such that for every positive integer $n \ge n_m$, there exist positive integers $a_1, a_2, \ldots, a_n$ such that$$\frac{1}{a_1^m}+\frac{1}{a_2^m}+\ldots+\frac{1}{a_n^m}=1.$$
''Solution:'':
The "grand" idea of this question is to use Bèzout's identity. So note that $\sum_{i=1}^{k^m}\frac{1}{(ak)^m}=\frac{1}{a^m}$. Then, in fact, if $a=1$, we can write $1$ as the sum of $k^m$ numbers of the form $\frac{1}{k^m}$:
$1=\sum_{i=1}^{k^m}\frac{1}{(k)^m}$, with $n_0=k^m$.
Now, using the last idea, we can write $\frac{1}{k^m}$ as $\sum_{i=1}^{p^m}\frac{1}{(pk)^m}$ (1), adding $p^m -1$ terms to this sum or write $\frac{1}{k^m}$ as $\sum_{i=1}^{q^m}\frac{1}{(qk)^m} $(2), adding $q^m -1$ terms to the sum. Then, the resulting $n$ is equal to: $k^m +(p^m-1)u +(q^m-1)v$, after replacing (1) $u$ times and (2) $v$ times.
We only need to have $\gcd(p^m-1,q^m-1)=1$, because for $\gcd(A,B)=1$ and $a,b$ varying on non-negative integers, then $n= aA+bB$ always has a solution if $n>AB-A-B$ [Chicken McNugget Theorem].
Then we just need to have $\gcd(p^m - 1,q^m-1)=1$. Take $q=l(p^m -1)$ for some positive integer $l$ and it is over, as desired.
Correct?
If correct, would you have an easier way to prove it?
What you've done is basically correct except your proof is not quite complete. The issue is it requires $u + v \le k^m$. However, no matter how large $k$ is, as $n$ gets large enough, this inequality will not always hold. Nonetheless, it's relatively easy & simple to handle this problem.
First, as you noted about the coin problem (with the Chicken McNugget Theorem being a special case, which is applicable here, of where there are $2$ terms), for any choice of integers $p \gt 1$ and $q \gt 1$ so $\gcd(p^m - 1, q^m - 1) = 1$, set
$$A = p^m - 1, \; B = q^m - 1, \; j = AB - A - B \tag{1}\label{eq1A}$$
With any integer $k_0$, have
$$k_i = k_0 + i \; \forall \; i \ge 0 \tag{2}\label{eq2A}$$
Next, set $i = 0$ and choose a $k_0$ large enough so, for each
$$k_i^m + j + 1 \le n \le k_{i+1}^m + j \tag{3}\label{eq3A}$$
there exist non-negative integers $u$ and $v$ where
$$n = k_i^m + (p^m - 1)u + (q^m - 1)v \tag{4}\label{eq4A}$$
and
$$u + v \le k_i^m \tag{5}\label{eq5A}$$
As an exercise for you to do, show for all $i \gt 0$ that, for each $n$ satisfying \eqref{eq3A}, there are non-negative integers $u$ and $v$ so \eqref{eq4A} and \eqref{eq5A} are both true.
This proves, with $n_m = k_0^m + j + 1$, that for all $n \ge n_m$ there exists positive integers $a_i \; \forall \; 1 \le i \le n$ such that
$$\frac{1}{a_1^m} + \frac{1}{a_2^m} + \ldots + \frac{1}{a_n^m} = 1 \tag{6}\label{eq6A}$$
As for there being any easier way to prove \eqref{eq6A}, I don't know of any offhand.