Under what situations does $x+1$ divide $4n^2-x$? I am looking at the equation 
$$\frac{4n^2-x}{x+1} = y$$
for even $x$ and $y$, both positive. Under what situations does $x+1$ divide $4n^2-x$?

I will throw another wooden nickel on the fire.
Since $\dfrac{4n^2-x}{x+1}+1=\dfrac{4n^2+1}{x+1}$, we get that the question is equivalent to asking when
$$
x+1\,|\,4n^2+1\tag{1}
$$
Trivially, $x=0$ and $x=4n^2$ satisfy $(1)$, and these are the only solutions when $4n^2+1$ is prime. Let's look for less trivial solutions. That is, let us look at when $4n^2+1$ is composite.
Suppose there is a prime $p$ so that
$$
p\,|\,4n^2+1\tag{2}
$$
$4n^2+1$ is odd, so $p$ must be odd. Furthermore, condition $(2)$ says that $-1\equiv(2n)^2\pmod{p}$; that is, $-1$ is the square of some number $\bmod{p}$ (usually said as "$-1$ is a quadratic residue $\bmod{p}$").

Intro to Quadratic Residues
Suppose that $p$ is an odd prime and $a\not\equiv0\pmod{p}$ is a quadratic residue $\bmod{p}$. There is an $x$ so that $x^2\equiv a\pmod{p}$. First, note that $x\not\equiv0\pmod{p}$. Therefore, by Fermat's Little Theorem,
$$
a^{\frac{p-1}{2}}\equiv x^{p-1}\equiv1\pmod{p}\tag{3}
$$
Consider the equation
$$
(x^{\frac{p-1}{2}}+1)(x^{\frac{p-1}{2}}-1)=x^{p-1}-1\equiv0\pmod{p}\tag{4}
$$
Since $\mathbb{Z}_p$ is a field, there are at most $\frac{p-1}{2}$ solutions to each of
$$
x^{\frac{p-1}{2}}+1\equiv0\pmod{p}\quad\text{and}\quad x^{\frac{p-1}{2}}-1\equiv0\pmod{p}\tag{5}
$$ 
However, Fermat's Little Theorem says there are $p-1$ solutions to $(4)$. Therefore, there are exactly $\frac{p-1}{2}$ solutions to each relation in $(5)$.
Consider the set of non-zero, quadratic residue classes, $Q$. For each $a\in Q$, there are exactly two solutions for $x^2=a$ ($x$ and $-x$). Thus, $|Q|=\frac{p-1}{2}$. For each $a\in Q$, we showed in $(3)$ that $a^{\frac{p-1}{2}}\equiv1$. The pigeonhole principle and $(5)$ say that for the $\frac{p-1}{2}$ non-zero, non-quadratic residue classes $\bmod{p}$, we have $a^{\frac{p-1}{2}}\equiv-1\pmod{p}$.
Thus, we have shown that for an odd prime $p$,
$$
a^{\frac{p-1}{2}}\equiv\left\{\begin{array}{}+1\pmod{p}&\text{if }a\text{ is a quadratic residue }\bmod{p}\\-1\pmod{p}&\text{if }a\text{ is not a quadratic residue }\bmod{p}\end{array}\right.\tag{6}
$$
For example, $-1$ is a quadratic residue $\bmod{p}$ if and only if $(-1)^{\frac{p-1}{2}}\equiv1\pmod{p}$; that is, if and only if $p\equiv1\pmod{4}$.

Therefore, if $x+1$ is divisible by any prime $p\not\equiv1\pmod{4}$, then $\dfrac{4n^2-x}{x+1}$ cannot be an integer.
If $4n^2+1$ is prime, there are only trivial solutions, and if $4n^2+1$ is composite, then $x+1$ can be any factor of $4n^2+1$.
