Has this conjecture been discovered before? (Concerns prime numbers and not $\#(n)$ like in a question similar to this)

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This is my second question today (I want to keep my major questions seperate):

Let's denote the $n^{th}$ prime number as $p_n$. Given the equation:

$$\frac{p_1p_2p_3 \cdots p_n}{1 + p_1 + p_2 + p_3 + \cdots + p_n} = \{x_m : x_m \in\mathbb{Z}^+\}$$

Now the $LHS$ is not always equal to a positive integer, but when it is, it is equal to $x_m$. The first positive integer it will be equal to (which is $1$) is denoted as $x_1$ such that $m = 1$ and $n = 2$, and the second integer (which is $715$) is denoted as $x_2$ such that $m = 2$ and $n = 6$ etc.

Conjecture:

$$δ(x_m) = \{2^{2m - 1} : m > 1 \ \land \ δ(x_m) = \text{number of divisors of $x_m$ including $x_m$ itself}\}$$

$$m > 1 \because δ(x_1) = δ(1) = 1 \neq 2^{2\times 1 - 1} = 2$$

Is there a way to prove/disprove this? This is my own conjecture and I put my own little symbols to it. Could somebody please explain to me why this works and any examples where it doesn't (if they exist)? If this is somebody else's conjecture then I will delete this post. I am in Year $9$, so my skill level in mathematics is not too high (compared to most users on this site), so I am sure somebody else must have discovered this.

Thanks to this site, the only non-Year $9$ maths I am familiar with include:

  • Sets
  • Matrices
  • Sum/product series Σ and Π
  • Imaginary (lateral) numbers $\sqrt{\text{negative}}$ (not particularly complex numbers $a + bi$)
  • Quadratic equations
  • Logarithms $\log_bn = m$
  • Trigonometry $\Delta ABC$ and
  • Linear Graphs $y = mx + c$.

I guess these are the only things that may make you wonder if I actually am in Year $9$. Everything else that what a Year $9$ is usually taught, I also know. If you are using other techniques and ways of proving/disproving my conjecture, please be clear and try not to skip too many steps. I have not decided to name my conjecture as of yet (and I don't want to be selfish by naming it based on my own name) but I want to see what you think of it.

Thank you in advance.