I am unable to find the general closed form and would appreciate if you can help me. Thank you.
$k\ge0$
$$\sum_{n=0}^{\infty}{2^n(n^{2k+1}+n^{2k}+n^{2k-1}+\cdots+ny+x)\over (n+1)(2n+1)(2n+3){2n\choose n}}=F(k)$$
$$\sum_{n=0}^{\infty}{2^n(ny+x)\over (n+1)(2n+1)(2n+3){2n\choose n}}=F(0)$$
$$\sum_{n=0}^{\infty}{2^n(n^3+n^2+ny+x)\over (n+1)(2n+1)(2n+3){2n\choose n}}=F(1)$$
$$\sum_{n=0}^{\infty}{2^n(n^5+n^4+n^3+n^2+ny+x)\over (n+1)(2n+1)(2n+3){2n\choose n}}=F(2)$$
I only managed to work out the closed form for $k=0,1$ and $2$
$$F(0)=(x-y)\cdot{\pi^2\over 8}+(2x-3y)\cdot{\pi\over 2}-(4x-6y)$$
$$F(1)=(x-y)\cdot{\pi^2\over 8}+(2x-3y-3)\cdot{\pi\over 2}-(4x-6y-5)$$
$$F(2)=(x-y)\cdot{\pi^2\over 8}+(2x-3y-9)\cdot{\pi\over 2}-(4x-6y-18)$$
I found these using some algebra manipulation and trial and error via a sum calculator.
Partial Solution
@HazemOrabi already covered most of this in his comment, but I'll try to expand the idea in his comment into a bit more into an answer.
You wrote
However, you are abusing summation notation in the numerator. You can't use middle dots to imply a continued sum when the last two terms are different. Also, the variable $k$ shows up in your sum for no reason. What I presume you mean (from reading your closed forms for fixed $M$) is that $$F(M)=\sum_{n=0}^{\infty}{2^n[ny+x]+2^n[(n^{2M+1}+n^{2M})+\cdots+(n^5+n^4)+(n^3+n^2)]\over (n+1)(2n+1)(2n+3){2n\choose n}}$$ We can rewrite the numerator as $$F(M)=\sum_{n=0}^{\infty}{2^n[ny+x]+2^n\sum_{k=1}^M (n^{2k}+n^{2k+1})\over (n+1)(2n+1)(2n+3){2n\choose n}}$$ We can split the two terms in the numerator, and we know that the first converges and is equal to $F(0)$ $$F(M)=\sum_{n=0}^{\infty}{2^n[ny+x] \over (n+1)(2n+1)(2n+3){2n\choose n}} +\sum_{n=0}^{\infty}{2^n\sum_{k=1}^M (n^{2k}+n^{2k+1})\over (n+1)(2n+1)(2n+3){2n\choose n}}$$ $$F(M)=F(0)+\sum_{n=0}^{\infty}{2^n\sum_{k=1}^M (n^{2k}+n^{2k+1})\over (n+1)(2n+1)(2n+3){2n\choose n}}$$ When $n=0$ the sum is zero, so we change index $$F(M)=F(0)+\sum_{n=\color{red}{1}}^{\infty}{2^n\sum_{k=1}^M (n^{2k}+n^{2k+1})\over (n+1)(2n+1)(2n+3){2n\choose n}}$$ We now bring the $(n+1)$ from the denominator into the sum in the numerator and do some canceling, noting that $n^{2k} + n^{2k+1} = n^{2k}(n+1)$ $$F(M)=F(0)+\sum_{n=1}^{\infty}{2^n\sum_{k=1}^M (\frac{n^{2k}+n^{2k+1}}{n+1})\over (2n+1)(2n+3){2n\choose n}}=F(0)+\sum_{n=1}^{\infty}{2^n\sum_{k=1}^M n^{2k}\over (2n+1)(2n+3){2n\choose n}}$$ We can now sum in a different order and thus pull the inner summation in the numerator to the outside. If you don't trust me, try expanding the sum and grouping in different ways to get this result $$F(M)=F(0)+\sum_{k=1}^M\sum_{n=1}^{\infty}{2^nn^{2k}\over (2n+1)(2n+3){2n\choose n}}$$ We now split the fraction into two using Partial Fraction Decomposition on $\frac{1}{(2n+1)(2n+3)}$ $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}}{{2n\choose n}}\left[\frac{1}{(2n+1)}-\frac{1}{(2n+3)}\right]$$ We now distribute and split the inner summation to get $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}}{{2n\choose n}}\left[\frac{1}{(2n+1)}-\frac{1}{(2n+3)}\right]$$ $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}}{{2n\choose n}(2n+1)}-\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}}{{2n\choose n}(2n+3)}\right)$$ We now use the fact that ${2n \choose 2} = \frac{(2n)!}{(n!)^2}$ $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}(n!)^2}{(2n)!(2n+1)}-\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}(n!)^2}{(2n)!(2n+3)}\right)$$ $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}(n!)^2}{(2n)!(2n+1)}-\sum_{n=1}^{\infty}\frac{2^n\,n^{2k}(n!)^2}{(2n)!(2n+3)}\right)$$ We now use the fact that $(2n-1)!! = \frac{(2n)!}{2^n n!}$ $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}}{(2n-1)!!\,(2n+1)}-\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}}{(2n-1)!!\,(2n+3)}\right)$$ We can simplify the denominators a bit $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}}{(2n+1)!!}-\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}\,(2n+1)}{(2n+3)!!\,}\right)$$ We can simplify the denominators a bit $$F(M)=F(0)+\frac{1}{2}\sum_{k=1}^M\left(\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}}{(2n+1)!!}-\sum_{n=1}^{\infty}\frac{(n!)\,n^{2k}\,(2n+1)}{(2n+3)!!\,}\right)$$ This appears similar to $$\arcsin(x)=\sum_{n=0}^\infty \frac{(2n)!}{2^{2n}(n!)^2(2n+1)}x^{2n+1}=\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!(2n+1)}x^{2n+1}$$ But I can't figure out how to solve from here