You have the basic right idea of showing the $\operatorname{lcm}$ values increase faster than the cube of the highest value used in the $\operatorname{lcm}$, with the following being one way to finish the solution. Define for any positive integer $m$,
$$f(m) = \operatorname{lcm}(1,2,\ldots,m) \tag{1}\label{eq1A}$$
For some prime $m \gt 8$, consider if you have
$$f(m) \gt 8m^3 \tag{2}\label{eq2A}$$
For any integer $n \ge m^3$, since $\sqrt[3]{n} \ge m$, you would need to include $m$ in the $\operatorname{lcm}$. However, \eqref{eq2A} shows you actually need $n \gt 8m^3$, so $\sqrt[3]{n} \gt 2m$. The less restrictive formulation of Betrand's postulate states there's always at least one prime $p$ where $m \lt p \lt 2m$, so since $p \gt 8$ and this prime $p$ must be multiplied into the $\operatorname{lcm}$ value, you have
$$f(2m) \gt p(8m^3) \gt 8(8m^3) = 64m^3 \tag{3}\label{eq3A}$$
Thus, you actually have $n \gt 64m^3$, giving $\sqrt[3]{n} \gt 4m$. You can use the procedure above repeatedly, in an inductive type fashion, to get $n \gt (8^{k})m^{3}$ for $k = 1, 2, 3, \ldots$, showing there is no larger $n$ which works.
As for the base case, note that
$$f(11) = 27\text{,}720 \gt 10\text{,}648 = 8(11^3) \tag{4}\label{eq4A}$$
Since it seems you've checked the other cases for $m \lt 11$, this shows the largest $n$ which works is what you've already found, i.e., $420$.
This answer is based on the excellent work of Will Jagy. This solves all cases of $k>3.$
Let $p<k$ be an odd prime such that $p\not\mid k.$
Solve $kd\equiv -1\pmod{p}.$ Let $n=(kd+1)/p.$ Note that since $p<k,$ $n>d.$
Then for any integer $t,$ we can take $z=a^{d}t^p$ so that $$\begin{align}az^k+1&=a^{kd+1}t^{kp}+1\\&=\left(a^nt^k\right)^p+1\\
&=(a^nt^k+1)\left(1+a^nt^k\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}\right)
\end{align}$$
Where the last equation is because when $p$ is odd, $$
\begin{align}u^p+1&=(u+1)
\sum_{j=0}^{p-1} (-1)^ju^j
\\&=(u+1)\left(1+u\sum_{j=1}^{p-1}(-1)^ju^{j-1}\right)\end{align}$$
Now, since $n>d,$ we can set $$
\begin{align}x&=a^{n-d}t^{k-p}\\
y&=a^{n-d}t^{k-p}\sum_{j=1}^{p-1} (-1)^j\left(a^nt^k\right)^{j-1}
\end{align}$$
For $k\geq 4$ we can always find such a $p$ by taking a prime factor of $n-1$ or $n-2$ if $n$ is even or odd, respectively.
So this solves all cases $k>3.$
You don't need $p$ prime, just that $1<p<k$ is odd and $\gcd(p,k)=1.$
k even
So when $k$ is even, we can take $p=k-1.$ Then $d=p-1$ and $n=p.$
Then for any integer $t,$ $$\begin{align}z&=a^{k-2}t^{k-1}\\x&=at\\y&=at\sum_{j=1}^{k-2}(-1)^j\left(a^{k-1}t^k\right)^{j-1}.\end{align}$$
k odd
Likewise, if $k=2m+1$ is odd, then you can take $p=2m-1,$ $d=m-1$ and $n=m.$ Then for any integer $t$:
$$\begin{align}z&=a^{m-1}t^{2m-1}\\
x&=at^2\\
y&=at^2\sum_{j=1}^{2m-2}(-1)^j\left(a^mt^{2m+1}\right)^{j-1}
\end{align}$$
is a solution.
In particular, for $k>3$ there are infinitely many solutions $(x,y,z)$ with $a\mid x$ and $x\mid y$ and $x\mid z.$
Best Answer
If you focus on just finding an $n$ with many solutions (obviously $2022$ in the problem statement is just a placeholder for 'how many you want'), this problem is easy if you go the other way around: Find parameterized solutions $(x,y,n)$, then choose the parameters such that the same $n$ appears as often as you want.
Since $\sqrt{xy}$ needs to be an integer, given any $x$, choosing $y=x$ seems to be the most immediate choice to make that true. That means $n=x+x+\sqrt{x\times x}=3x$. This means $(x,x,3x)$ is a solution triple.
That means any $n$ divisible by $3$ has at least $(x,y)$ solution, namely $x=y=\frac{n}3$. But for a given $n$, this is only one solution.
But the method can be easily generalized by seeing that $y=k^2x$ for $k \in \mathbb N, k \ge 1$ also leads to a natural $\sqrt{xy}$ (the above is just $k=1$).
In that case, we get $n=x+k^2x+\sqrt{x\times k^2x} = x+k^2x+kx=(1+k+k^2)x.$
That means for each positive $x$ and $k \in \mathbb N$,
$$(x, k^2x,(1+k+k^2)x) \tag{1}$$ is a solution triple.
That means any $n$ divisible by $1+k+k^2$ has a solution of the form $(1)$. That means you 'only' need to find an $n$ that is divisible by $1+k+k^2$ for $2022$ different values of $k$, say $k=1,2,\ldots,2022$. It will be very large, but it certainly exists, let's call it $n_{2022}$.
For different $k$ the $x$ and $y$ values generated by $(1)$ are all different, because for example $x=\frac{n_{2022}}{1+k+k^2}$ and the denominator is a strictly increasing function of $k$ for positive $k$, so all the $x$ values are different.