Direct Method
Putting $y_n=(n+1)!2^{1-n}x_n$ the recurrence relation
$$
y_{n+1}−\frac{n+2}{2}y_n=(n+1)(n+2)3^n\tag 1
$$
with initial condition $y_0=0$ forall $n\ge 0$, becomes
$$
x_{n+1}=x_n+\frac{6^n}{n!}. \tag 2
$$
with initial condition $x_0=\frac{y_0}{2}=0$.
This is a non-homogeneous recurrence relation with non-homogeneous term $\xi_n=\frac{6^n}{n!}$ that can be solved directly in a very simple way.
We know that the general solution to a non-homogeneous recurrence relation is the sum of the general solution to the associated homogeneous recurrence and any particular solution to the non-homogeneous recurence. So if $a_n$ is a solution to the associated homogeneous recurrence $x_{n+1}-x_n=0$, and $b_n$ is a particular solution to the non-homogeneous recurrence $x_{n+1}-x_n=\xi_n$, then $a_n+b_n$ is the general solution to the non-homogeneous recurrence.
The general solution $a_n$ of the associated homogeneous recurrence relation $x_{n+1}-x_n=0$ is trivial, i.e. $a_n=0$ for all $n\ge0$. Infact the characteristic equation is $\lambda-1=0$, so that $a_n=\alpha^1$ and from the initial condition $x_0=0$, we have $\alpha=0$, i.e $a_n=0$.
Now we have to find a particular solution $x_n=b_n$ of the relation (2). Observing that the term $\xi_n=\frac{6^n}{n!}$ is the $n$-th term of the exponential series, and that the difference between the $n+1$-th term $b_{n+1}$ and the $n$-th term $b_n$ must be the $n$-th term of the exponential series, we may take a particular solution as a truncated exponential series
$$
b_n=\sum_{n=0}^{n-1}\frac{6^k}{k!}.
$$
It's evident that $b_n$ satisfies the relation(2):
$$
\begin{align}
b_{n+1}-b_n&=\sum_{k=0}^{n}\frac{6^k}{k!}-\sum_{k=0}^{n-1}\frac{6^k}{k!}\\
&=\left(1+\tfrac{6^1}{1!}+\cdots+\tfrac{6^{n-1}}{(n-1)!}+\frac{6^{n}}{n!}\right)-\left(1+\tfrac{6^1}{1!}+\cdots+\tfrac{6^{n-1}}{(n-1)!}\right)\\
&=\frac{6^{n}}{n!}.
\end{align}
$$
So the general solution of (2) is
$$
x_n=a_n+b_n=\sum_{k=0}^{n-1}\frac{6^k}{k!}.\tag 3
$$
Finally the general solution of (1) is
$$
y_n=(n+1)!2^{1-n}\sum_{k=0}^{n-1}\frac{6^k}{k!}\tag 4
$$
for $n\ge 1$ and $y_0=0$.
Ordinary Generating Function
If you prefer to work with generating function, multiply the recurrence (2) by $z^n$ and sum over $n$
$$
\sum_{n=0}^\infty x_{n+1}z^n-\sum_{n=0}^\infty x_{n}z^n=\sum_{n=0}^\infty \frac{6^n}{n!}z^n
$$
and put $X(z)=\sum_{n=0}^{\infty} x_nz^n$ for $|z|<1$; we have
$$
\frac{1}{z}(X(z)-x_0)-X(z)=\operatorname{e}^{6z}
$$
that is
$$
X(z)=\frac{z}{1-z}\cdot \operatorname{e}^{6z}.
$$
Observing that $\frac{z}{1-z}=\sum_{n=1}^{\infty}z^n$, then $x_n$ is the discrete convolution of the discrete Heaviside step function $h_n$ and the sequence $\xi_n=\frac{6^n}{n!}$, that is
$$x_n=\sum_{k=0}^{n-1}\frac{6^k}{k!}$$
Other Representations
Recalling that the incomplete gamma function $\Gamma(\alpha,x)$ is given by
$$
\Gamma(\alpha,x)=\int_x^\infty t^{\alpha-1}\operatorname{e}^{-t}\operatorname{d}t
$$
and that for $\alpha$ an integer $n$
$$
\Gamma(n,x)=(n-1)!\operatorname{e}^{-x}\sum_{k=0}^{n-1}\frac{x^k}{k!}=(n-1)!\operatorname{e}^{-x}e_{n-1}(x),
$$
where $e_n(x)=\displaystyle\sum_{k=0}^{n-1}\frac{x^k}{k!}$ is the exponential sum function, the general solution (4) can be expressed as
$$
y_n=(n+1)!2^{1-n}e_{n-1}(6)
=\operatorname{e}^{6}2^{1-n}(n+1)n\Gamma(n,6)\tag 5
$$
Using the Generalized Exponential Integral $E_n$ function defined as
$$
E_n(x)= \int_1^\infty\frac{\operatorname{e}^{-xt}}{t^n}\operatorname{d}t= x^{n-1}\Gamma(1-n,x)
$$
the solution (5) can be expressed as
$$
y_n=\operatorname{e}^{6}2^{1-n}(n+1)n6^nE_{1-n}(6).\tag 6
$$
Finally, putting all together the general solution of (1) can be represented as
$$
y_n=\operatorname{e}^{6}2^{1-n}(n+1)n\Gamma(n,6)=\operatorname{e}^{6}2^{1-n}(n+1)n6^nE_{1-n}(6).\tag 7
$$
