Prime exponent Diophantine equations with infinitely many solutions

The question can be simplified down to: does there exists naturals $a_i$ for $1\leq i \leq n$ such that

$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$

This is in fact true. Let $L=p_1p_2\cdots p_{n-1}$. Clearly, $p_i|L$ for $1\leq i\leq n-1$ and that $\gcd\left(p_n,L\right)=1$. Thus, there exists $x,y\in\mathbb{N}$ such that

$$xp_n-yL=1\Rightarrow yL=xp_n-1$$

Now, for $1\leq i\leq n-1$, define

$$a_i=y\frac{L}{p_i}$$

and

$$a_n=x$$

Then

$$a_ip_i=yL\text{ for }1\leq i \leq n-1$$

and

$$a_np_n=xp_n=yL+1$$

Having proved there exists naturals such that

$$a_1p_1=a_2p_2=\dots=a_{n-1}p_{n-1}=a_np_n-1$$

is true, we move on to the end of the proof. As @John_Omiela noted in his comment, if you have one solution you have infinite solutions. One solution is

$$x_i=(n-1)^{a_i}$$

Then

$$x_1^{p_1}+x_2^{p_2}+\cdots +x_{n-1}^{p_{n-1}}=(n-1)^{a_1p_1}+(n-1)^{a_2p_2}+\cdots+(n-1)^{a_{n-1}p_{n-1}}$$

$$=(n-1)(n-1)^{yL}=(n-1)^{yL+1}$$

while

$$x_n^{p_n}=(n-1)^{a_np_n}=(n-1)^{yL+1}$$

Thus, equality has been established and we are done.

Yes, you can always have an infinite number solutions solutions. First, let

$$Q = \prod_{i=1}^{n-1}p_i \tag{1}\label{eq1A}$$

Next, since $Q$ would be relatively prime to $p_n$, then you have

$$kQ \equiv -1 \pmod{p_n} \implies k \equiv -Q^{-1} \pmod{p_n} \tag{2}\label{eq2A}$$

Choose one of these $k$ and then have

$$x_i = \left(n-1\right)^{\frac{kQ}{p_i}}, \; 1 \le i \le n - 1 \tag{3}\label{eq3A}$$

The equation you're asking to solve is

$$\sum_{i=1}^{n-1}x_i^{p_i} = x_n^{p_n} \tag{4}\label{eq4A}$$

The LHS side of \eqref{eq4A} is then

$$\begin{equation}\begin{aligned} \sum_{i=1}^{n-1}x_i^{p_i} & = \sum_{i=1}^{n-1}\left(\left(n-1\right)^{\frac{kQ}{p_i}}\right)^{p_i} \\ & = \sum_{i=1}^{n-1}\left(n-1\right)^{kQ} \\ & = \left(n-1\right)^{kQ+1} \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

However, by \eqref{eq2A}, $kQ + 1$ is a multiple of $p_n$, say

$$kQ + 1 = mp_n \tag{6}\label{eq6A}$$

Thus, you can set

$$x_n = (n-1)^{m} \tag{7}\label{eq7A}$$

This shows a set of values which solve \eqref{eq4A}.

As for getting an infinite number of solutions, have

$$R = \prod_{i=1}^{n}p_i \tag{8}\label{eq8A}$$

Next, choose any

$$m \gt 1, \; m \in \mathbb{N} \tag{9}\label{eq9A}$$

Then consider

$$x_i^{'} = \left(m^{\frac{R}{p_i}}\right)x_i, \; 1 \le i \le n \tag{10}\label{eq10A}$$

Each term on the left & right of \eqref{eq4A} would be multiplied by $m^{R}$, so it would still be satisfied.