I came up with this (I admit I'm probably not the first one to have this thought but I haven't been able to find anyone else with the same question) while reading about semiprimes.
Clearly $y$ is always odd which suggests to me that it could also be prime a lot of the time. Additionally, There are infinitely many semiprimes so it seems somewhat likely (for purely intuitive reasons) that for semiprime $n$ it could be that $2n+1$ could also be prime.
Comment: Some experiments:
We know all primes are of the following forms:
$30k+r; r=1,7, 11, 13, 17, 19, 23, 29$
with some conditions we can construct primes of the form $y=2p_1p_2+1$, for example:
$p_1=30k+1$, $p_2=30k+11$, for $k=1$ gives $y=2543$
$p_1=3ok+11$, $p_2=30k+13$, for $k=1$ gives $y=3527$
some conditions for $k=1$ are:
$\begin{cases}r_{p_1}\not\equiv 3\bmod 10\\r_{p_1}\not\equiv 9\bmod 10\end{cases}$
because:
$y\equiv (2\cdot 3\cdot 9+1=55)=0\bmod 5$
That is y will be divisible by 5.
Or $p_1= 30k+11$, $p_2=30k+23$ for $k=1$ gives $y=4347=3\cdot 1449$
Now suppose there exist some $k$ for which we can construct $y$ taking primes of the form $p=3ok+r$, then we have at least $n$ primes , where:
$n={8\choose 2}=28$
We do not know what magnitudes of $r_{p_1}$ and $r_{p_2}$ with how many $k$ make primes $y$, but there is no limit for our choices of these three parameters.
$p_1$ and $p_2$ must not satisfy following identity:
$(5x+13)^2+1=(3x+7)^2+(4x+11)^2$
$p_1\neq(3x+7)$ and $p_2\neq(4x+1)$
because:
$(5x+13)^2+1=(3x+7)^2+(4x+11)^2$
$\Rightarrow (5x+13)^2+1+2(3x+7)(4x+11)=(3x+7+4x+11)^2$
$\Rightarrow 1+2p_1p_2=(7x+18)^2-(5x+13)^2=(2x+5)(12x+31)$
which is composite.