Inequality with binomial coefficients $ \binom{2n-1}{n} + \binom{2n-1}{n+1} + \cdots + \binom{2n-1}{n+z} \geq (2z+3)\binom{2n-2}{n+z} $

Determine all positive integers $n$ and $z \leq n-1$ such that $$ \binom{2n-1}{n} + \binom{2n-1}{n+1} + \cdots + \binom{2n-1}{n+z} \geq (2z+3)\binom{2n-2}{n+z} $$ I tried small $z$, e.g. $z=4$ and the inequality fails for large $n$. On the other hand, a crude approach such as the left-hand side being at least $(z+1)\binom{2n-1}{n+z}$ works if and only if $$ \frac{\binom{2n-1}{n+z}}{\binom{2n-2}{n+z}} \geq \frac{2z+3}{z+1} \Leftrightarrow \frac{2n-1}{n-z-1} \geq \frac{2z+3}{z+1} \Leftrightarrow 2(z+1)^2 \geq n $$ i.e. $z$ is of order $\sqrt{n}$ or higher. So it could be indeed the case that, given $n$, the possible $z$ are from something about $\sqrt{n}$ to $n-1$, but I have no idea how to even start proving this, as I cannot find any useful strong non-asymptotic inequalities anywhere.

Any help appreciated!

This isn't a full answer to your question (which I suspect is impossible to determine without explicit computation), or even a solid upper bound on the minimum $n$ for some $z$, but I do have a significantly better (approximate) approximation for the lower bound of n than you do which can be solidified with more analysis.

We start with a reformulation of the question. Divide both sides of the inequality by ${2n-2\choose n+z}$ and subtract $2z+3$ from both sides. We then get $f(n,z)=\sum_{j=0}^{z}\frac{{2n-1\choose n+j}}{{2n-2\choose n+z}}-(2z+3)$, and our goal is to find when $f(n,z)\geq 0$. Define $g(n,z,j)=\frac{{2n-1\choose n+j}}{{2n-2\choose n+z}}$ such that $f(n,z)=\sum_{j=0}^z g(n,z,j)-(2z+3)$. We can simplify $g$ to $\frac{2n-1}{n-j-1}\prod_{i=0}^{z-j-1}\left(1+\frac{z+j+2}{n-z-1+i}\right)$. From here, we can get our first bound for $f$, by bounded $g$ from above and below. We are primarily dealing with small $z$ and large $n$, so each term in the product is positive so the product is positive. This means to get a lower bound for $g$ we should try to minimize the term in the product, and to get an upper bound we maximize the term. The minimum term is when $i=z-j-1$ and the maximum term is when $i=0$, so we get the following upper and lower bounds for $g$: $$ \frac{2n-1}{n-j-1}\left(1+\frac{z+j+2}{n-j-2}\right)^{z-j}\leq g\leq \frac{2n-1}{n-j-1}\left(1+\frac{z+j+2}{n-z-1}\right)^{z-j} $$ which simplify to the bounds: $$ \frac{2n-1}{n-j-1}\left(\frac{n+z}{n-j-2}\right)^{z-j}\leq g\leq \frac{2n-1}{n-j-1}\left(\frac{n+j+1}{n-z-1}\right)^{z-j}. $$ These bounds happen to be incredibly tight, especially when $f$ is near $0$ or negative. We want to test the lower bound of $f$, such that $$0\leq \sum_{j=0}^{z}\frac{2n-1}{n-j-1}\left(\frac{n+z}{n-j-2}\right)^{z-j} - (2x+3)\leq f.$$ Now, the trick here is to notice that compared to $n$, $j$ is small beans. We then attempt to fit a quadratic function to $g$ which matches $g$ on $j\in\{0,\frac{z}{2},z\}$. (This turns out to be a very good approximation, graphically tested to about $z\approx 20$.) This quadratic function has the form $$g'=\left(1-3\frac{j}{z}+2\frac{j^2}{z^2}\right)g(z,0)+\left(4\frac{j}{z}-4\frac{j^2}{z^2}\right)g\left(z,\frac{z}{2}\right)+\left(-\frac{j}{z}+2\frac{j^2}{z^2}\right)g(z,z).$$ This function is an approximate lower bound for $g$. Then, because we know the sums of constants, integers, and squares, we can easily replace the $j$s in this expression to get an approximate lower bound for $f$. $$f'=\frac{z+1}{6z}\left((z+2)g(n,z,0)+(z+2)g(n,z,z)+4(z-1)g\left(n,z,\frac{z}{2}\right)\right)-(2z+3).$$ Initially $n,z$ were integers, but now we let them range over the reals. One nice property of $f'$ is that some of the error as $g'$ approximates the lower bound for $g$ is averaged out, so $f'$ forms a proportionally better approximation for the lower bound of $f$. $f'$ also has the convenient properties that the limit of $f'$ as $n$ goes to infinity converges to a function of $z$, but if $z$ is allowed to range to infinity (potentially with $n$) then $f'$ is unbounded. Let $h(n,\epsilon)=f'\left(n,n^{\frac{1}{3}}\epsilon\right)$. Let $z'(n)$ be defined such that $f(n,z'(n))=0$. I claim that $z'(n)$ grows as fast as $n^{\frac{1}{3}}$. This means that for large values of $n$, there exists some constant $\epsilon$ such that $h(n,\epsilon)=0$. Suppose that $z'(n)$ grew slower than $n^{\frac{1}{3}}$, say $n^{\frac{1}{4}}$. If that were the case, then we would expect that the $\epsilon$ corresponding to $h(n,\epsilon)=0$ would "grow" as $n^{\frac{1}{4}-\frac{1}{3}}$ (which goes to $0$), and that for any constant $\epsilon$ the function $h(n,\epsilon)$ grows without bound. So, if the limit of $h(n,\epsilon)$ as $n$ goes to infinity converges to some function of $\epsilon$, then we've proven that $z'(n)$ grows like $n^{\frac{1}{3}}$.

For this part I just plugged it into sympy. The limit of $h$ as $n$ goes to infinity is $\frac{4}{3}\epsilon^3-1$, which means that in the limit, $\epsilon = \frac{3}{4}^\frac{1}{3}\implies h=0$. This means that $z'(n)$ probably grows no faster than $\left(\frac{3}{4}n\right)^\frac{1}{3}$. Or, in other words, a probable lower bound of $n$ is $n\geq \frac{4}{3}z^3$. I also ran my sympy script on the upper bound, and it gave me the same answer, which is strong evidence that the maximum lower bound of $n$ grows as $\frac{4}{3}z^3$.

I also did explicit computations (using that simplified form of $g$ to calculate $f$) to find the maximum lower bound of $n$ for different values of $z$:

z    1    2    3    4    5    6    7    8    9    10   11   12   13
n    14   41   92   177  302  476  707  1003 1372 1821 2359 2994 3734

I'll post my sympy code below so you can check it if you want:

from sympy import *
n,z,j,eps = symbols("n z j epsilon")
gl = (2*n-1)/(n-j-1)*((n+z)/(n-j-2))**(z-j)
gu = (2*n-1)/(n-j-1)*((n+j+1)/(n-z-1))**(z-j)
g = gl
# g = gu
f = (z+1)*((z+2)*g.subs(j,0) + (z+2)*g.subs(j,z) + 4*(z-1)*g.subs(j,z/2))/(6*z)-(2*z+3)
third = z/z/3
h = f.subs(z,n**third * eps)
# Proof that n grows like 4/3*x^3
print(limit(h,n,oo))