I love the answer of this question Also great title.

Suppose that $n$ is odd. $2d_2+d_4+d_5$ is even and $d_7$ is not even. Thus $n$ is even. If $3|n$, its clear that $d_3=3$, by second and thirth condition $d_7^2-d_6d_7+d_6^2=1$, contradiction. So $d_3>3$. If $4|n,d_3=4$ so $d_7=d_6+1$. And by second condition $8$ is a divisor too. If $5,7|n$ by firs t condition we get $16=d_7=d_6+1$ and $3|d_6$, if $5$ is not a divisor and $7$ is, we get $19=d_7$ and $3|d_6$, if $5|n$ and $7$ doesnt, $17=d_7$ and $n=1088$ but $5$ is not a divisor. if $5$ and $7$ both are not divisors, we get $d_5+12=d_7$ and $(d_5+12)(d_5+11)4=d_5d_{k-4}=n$ and we know that $d_5>7$ is a prime thus it must be $11$. $n=23\cdot 22\cdot 4=2024$. If $v_2(n)=1$Let $d_3=p$ If $d_6\geq 2p$. By Last two conditions we get $$d_6=\frac{d_7(p-2)-\sqrt{d_7^2(p^2-4p)+4}}{2}\geq 2p\Longrightarrow (d_7-4)p-2d_7\geq \sqrt{d_7^2(p^2-4p)+4}\Longrightarrow (d_7-4)^2p^2>d_7^2(p-3)^2+4$$gives a contradiction. Thus $d_6<2p$ so $d_4,d_5,d_6<2p$ are primes. And by $d_7$ is even its $2p$ thus we get $d_6=\frac{2p(p-2)-\sqrt{4p^2(p^2-4p)+4}}{2}$ and $\Delta=[2p(p-4)]^2-4\cdot 4p^2+4$ and it can't be perfect square eqsily. Thus theres no solution more. Thus $n=2024$.

It's so easy after saw that $d_7=d_6+1$.

The answer is only $\boxed{n = 2024}$. First we verify that $2024$ works. The divisors of $2024$ are $1,2,4,8,11,22,23,\dots$ which indeed satisfy $$\begin{aligned} 2 \cdot 2 + 8 + 11 &= 23 \\ 4 \cdot 22 \cdot 23 &= 2024 \\ (22 + 23)^2 &= 2024+1 \end{aligned}$$ Claim: $n$ is a multiple of $8$ Proof: First, notice that if \( n \) is odd, then all of its divisors will be odd, so \( 2d_2 + d_4 + d_5 \equiv 0 \not\equiv d_7 \pmod{2} \), a contradiction. Thus, \( n \) is even. Since \( n + 1 \) is an odd perfect square, it must satisfy \( n + 1 \equiv 1 \pmod{8} \), the claim follows. A corollary is that $d_2=2$. Claim: $d_3=4$ and $d_7=d_6+1$. Proof: Combining the last two equations, \[ (d_6-d_7)^2=n+1-\frac{4n}{d_3} \]If $d_3<4$ then the right hand side would be strictly smaller than $1$, a contradiction. Thus $d_3\geq 4$. As $4|n$ we must have that $d_3=4$. Thus $(d_6-d_7)^2=1$ and hence $d_7=d_6+1$. A consequence of the fact that $d_3=4$ is that $3\nmid n$. Substituting $d_2=2$ and $d_7=d_6+1$ into the first equation yields \[ 3 + d_4 + d_5 = d_6. \]We proceed with casework on $(d_4,d_5)$. Case 1: $d_4\neq 8$ (and thus a prime smaller than $8$) Case 1.1: $(d_4,d_5)=(5,7)$ Then $d_6=15$, skipping $8$. Case 1.2: $(d_4,d_5)=(5,8)$ Then $d_6=16$. But $5\nmid n=4\cdot 16\cdot 17$. Case 1.3: $(d_4,d_5)=(7,8)$ Then $d_6=18$, but $3\nmid n$. Case 2: $d_4=8$ Case 2.1: $d_5$ is even Then $d_5=16$, but then $d_6=27$, a multiple of $3$ Case 2.2: $d_5$ is odd Then $d_5|n=4d_6(d_6+1)=4(d_5+11)(d_5+12)$. As $d_5$ is the first odd divisor, it must be prime so either $d_5|11$ or $d_5|12$. The only possibility is $d_5=11$ giving $d_6=22$. Then $n=4\cdot 22\cdot 23=2024$, as desired.

To be honest this is just a nasty problem, which has potential mistakes lying hidden all over the place like landmines. We claim that the only answer is $n=2024$. We start with the following observation. Claim : We have $d_2=2$ and $d_3=4$. Proof : Say $n$ were odd then $d_2,d_4,d_5$ and $d_7$ are all odd. But then, \[d_7 = 2d_2+d_4+d_5 \equiv 0 + 1 + 1 \equiv 0 \pmod{2}\]which is a contradiction. Thus, $n$ is even which implies $d_2=2$. Further, since $n$ is even, \[(d_6+d_7)^2 = n+1 \equiv 1 \pmod{2}\]so $d_6+d_7$ is even. Since it is well known that the square of an odd integer is always $1\pmod{8}$ it follows that $n$ is divisible by $8$. Thus, $4\mid n$ so $d_3=3$ or $d_3=4$. Note that if $d_3=3$, \begin{align*} (d_6+d_7)^2 &= n+1 \\ d_6+2d_6d_7 + d_7^2 &= 3d_6d_7+1\\ d_6^2 + d_6^2 &= d_6d_7+1 \end{align*}But by AM-GM we have, $d_6^2+d_7^2 \ge 2d_6d_7 > d_6d_7 +1$ which is a clear contradiction. Thus, $d_3=4$ which proves the claim. Claim : The equality $d_7=d_6+1$ holds for any $n$ satisfying the given conditions. Proof : Note that since we established $d_3=4$, letting $d_7=d_6 + \epsilon$ we have \begin{align*} (d_6+d_7)^2 &= n+1 \\ (d_6+d_7)^2 &= 4d_6d_7 + 1\\ d_6^2 + d_7^2 &= 2d_6d_7 + 1\\ d_6^2 + (d_6+\epsilon)^2 &= 2d_6(d_6+\epsilon) +1\\ 2d_6^2 + 2d_6\epsilon + \epsilon^2 &= 2d_6^2 + 2d_6\epsilon +1\\ \epsilon^2 &= 1\\ \epsilon &= 1 \end{align*}which implies $d_7=d_6+1$ as claimed. Now note that since $(d_6+d_7)^2=n+1$ and the square of an integer is always $0$ or $1 \pmod{3}$, since $3\nmid n$ the only possibility is $3\mid n+1$ so $3\mid d_6+d_7=2d_6+1$ which implies $d_6 \equiv 1 \pmod{3}$. We now sadly have to resort to casework. Since $8\mid n$, $8$ is a factor of $n$. Further since $d_3=4$ , $d_8 \ge 9$. Thus, $8$ is one of $d_4$ , $d_5$ , $d_6$ and $d_7$. These will be our four cases. Case 1 : $d_7=8$. We then require $d_4=5$ , $d_5=6$ , $d_6=7$ and $d_7=8$, which is a contradiction since $3\nmid n$. Case 2 : $d_6=8$. But this is a contradiction since $d_6=8 \not \equiv 1 \pmod{3}$. Case 3 : $d_5=8$. Then we require, $d_4=5$ or $d_4=7$ (since $3\nmid n$ $d_4\ne 6$). If $d_4=5$, then $d_7=2d_2+d_4+d_5=17$ so $d_6=16$. This immediately gives $n=d_3d_6d_7=4\cdot 16 \cdot 17$ which is a contradiction since this is not divisible by $5$. If $d_4=7$, then we have $d_7=2d_2+d_4+d_5=19$ and $d_6=18$ which is a contradiction since $3\nmid n$. Case 4 : $d_4=8$. Here we have two possibilities. If $d_5=16$, then $d_7=2d_2+d_4+d_5 = 28$ and $d_6=27$ which is divisible by 3 which is impossible. If there exists some odd prime $p\mid d_5$, $d_7 = 2d_2+d_4+d_5=p+12$ so $d_6=p+11$. Since $d_5\mid n=d_3d_6d_7=4d_6d_7$. So $p\mid d_6$ or $p\mid d_7$. If $p\mid d_7$, $p\mid d_7-d_5=12$ which is impossible since $3\nmid n$. If $p \mid d_6$, $p\mid d_6-d_5 = 11$ so $p\mid 11$. Since no multiple of $11$ proceeds $d_5$ we must have $d_5=11$. So $d_6=22$ and $d_7=23$. Then, $n=4 \cdot 22 \cdot 23 = 2024$. It now remains to write out the divisors of $2024$ and check that all the conditions are met, which is easy enough.