Problem: Prove that there are infinitely many positive integer numbers that can't be written as
$4xy+3x+y$
OR
$4xy+3x-y$,
where $x,y$ are positive integers, and $y$ can be zero, too.
I tried to start with the following:
I set $4xy+3x+y=m$ and solved for $x$, so that $x=\frac{m-y}{4y+3}$ and as x is an integer, tried to find out the solution (also for $4xy+3x-y=m$), but got stucked.
Sorry, no solution but only some remarks.
Suppose the positive integer number $n$ cannot be written as either $4xy+3x+y$ or $4xy+3x-y$ for integers $x\geq1$ and $y\geq0$. Then the first restriction can be rewritten as $$n \neq 4 x y + 3 x + y$$ $$4 n + 3 \neq 16 x y + 12 x + 4 y + 3 = (4 x + 1)(4 y + 3)$$ Since any composite number of the form $4n+3$ can always be factorised like that, it follows by necessity that $4n+3$ needs to be prime number.
For the second restriction, we have $$n \neq 4 x y + 3 x - y$$ $$4 n - 3 \neq 16 x y + 12 x - 4 y - 3 = (4 x - 1)(4 y + 3)$$ which implies that the number $4n-3$ is not allowed to have any factors of the form $4 m+3$.
In particular, this would be true if both $4n-3$ and $4n+3$ would be prime, i.e. a sexy prime pair, but with the additional restriction that their average is a multiple of 4. Although it is conjectured that there are infinitely many sexy prime pairs, this is still unproven.
The first couple of numbers $n$ is given by $\{1,2,4,5,7,10,11,14,16,17,19,\dots \}$. Among these only 7 and 17 are not related to such as prime pair, i.e., for $n=7$ one would get $4 \times 7 \pm 3 = 25,31$.
In conclusion, any valid $n$ would imply that $4n+3$ is a prime and that $4n-3$ only contains prime factors of the form $4m+1$. To the best of my knowledge it has not yet been proven that there are infinitely many such numbers.