Consider the set of first $n$ primes. Distribute them into two disjoint sets. Then for each set, select each member at least once(maybe more than one) and get the product of selection to produce a number. Now you have two coprime numbers. Their difference does not contain any of the first $n$ prime as factor. Is it always possible to produce $(n+1)$st prime by choosing that two number close enough?
An example:
Consider sets ${[2,3,5]}$ and ${[7]}$.
$2*2*3*5 - 7*7 = 11$ gives the next prime.
Another set division of the same primes, $[2,5]$ and $[3,7]$
$3*7 - 2*5 = 11$ also gives the desired number.
It should, in principle, but appears to be impractical: I got$$13=2\cdot3^2\cdot5-7\cdot11=5\cdot7-2\cdot3\cdot11,$$$$17=2\cdot7\cdot13-3\cdot5\cdot11$$and$$19=2^2\cdot3\cdot5\cdot17-7\cdot11\cdot13,$$but got stumped at $23$. On the other hand, your approach can more easily generate other primes between $p_{n+1}$ and $p_{n+1}^2$ because a difference (or sum) of two coprime numbers can't be divisible by any of the numbers' factors; as examples, using the primes $2,3,5,7,11$, you can produce \begin{align}2\cdot3\cdot5^2-7\cdot11&=73,\\ 2^2\cdot7^2-3\cdot5\cdot11&=31,\\ 2\cdot3^3\cdot7-5^2\cdot11&=103,\\ 2\cdot3\cdot5\cdot7-11^2&=89,\\ 3\cdot5\cdot7\cdot11-2^{10}&=131,\\ 2^5\cdot11-3^2\cdot5\cdot7&=37,\\ 3\cdot5\cdot7-2^3\cdot11&=17, \end{align}as well as any difference (less than $13^2$) resulting from experimentation.
Some (perhaps all) can be written with fewer primes...the examples above, respectively, can be rewritten as \begin{align}73&=3\cdot5^2-2=2^4\cdot5-7;\\ 31&=3\cdot11-2=7^2-2\cdot3^2;\\ 103&=3\cdot5\cdot7-2=11^2-2\cdot3^2=5^3-2\cdot11;\\ 89&=2^2\cdot5^2-11=2^5\cdot3-7;\\ 131&=3\cdot7^2-2^4=3^3\cdot5-2^2=2^2\cdot5\cdot7-3^2;\\ 37&=2^4\cdot5-3=2^2\cdot11-7=7^2-2^2\cdot3;\\ 17&=5^2-2^3=3\cdot7-2^2=2\cdot11-5=2^2\cdot5-3. \end{align}In these cases, each representation found rules out a set of primes as divisors of its number; finding any with the excluded primes will lead to the number's primality.