Let $a:=\alpha$ and
\begin{equation*}
F_k(x,y):=\sum_{j=0}^\infty \frac{x^j}{j!}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y},
\end{equation*}
assuming the standard convention $0^0:=1$.
We have to consider the existence of a solution in $x$ and $y$ of the system
\begin{equation*}
a=F_k(x,y)=F_{k+1}(x,y). \tag{1}
\end{equation*}
We shall prove the following.

**Theorem 1.** Take any natural $k$ and any
\begin{equation*}
a\in(0,a_k],\quad\text{where}\quad a_k:=\sup_{x,y\ge0}F_{k+1}(x,y). \tag{1.5}
\end{equation*}
Then the system (1) has a solution $x,y\ge0$.

*Remark 1.* Since $F_k>0$, the condition $a\in(0,a_k]$ is obviously necessary in Theorem 1.

*Proof of Theorem 1.* Note that $F_k(x,y)\ge0$ for any real $x,y\ge0$ and $F_k(x,y)$ is continuous in real $x,y\ge0$. The crucial observation is the identity
\begin{equation*}
\partial_y F_{k+1}(x,y)=F_k(x,y)-F_{k+1}(x,y) \tag{2}
\end{equation*}
for real $x,y$.

Next, fix for a moment any real $x\ge0$. Then $F_{k+1}(x,0)=0$ and, by dominated convergence, $F_{k+1}(x,\infty-)=0$. So, $F_{k+1}(x,y)$ attains its maximum in $y$ at some real point $y=y_x\ge0$. At this point, we have $\partial_y F_{k+1}(x,y)=0$. So, by (2),
\begin{equation*}
F_k(x,y_x)=F_{k+1}(x,y_x)=\max_{y\ge0}F_{k+1}(x,y)=:M_{k+1}(x), \tag{3}
\end{equation*}
for all real $x\ge0$.

Next,
\begin{align*}
M_k(x)&\le\sum_{j=0}^\infty \frac{x^j}{j!}\max_{y\ge0}\frac{y^{k+j}}{(k+j)!}\,e^{-x-y} \\
&=\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac{(k+j)^{k+j}}{(k+j)!}\,e^{-k-j} \tag{4} \\
&\ll\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\frac1{\sqrt{k+j}}=E\frac1{\sqrt{k+\Pi_x}}
\underset{x\to\infty}\longrightarrow0
\end{align*}
by dominated convergence and because $\Pi_x\underset{x\to\infty}\longrightarrow\infty$ in probability, where $\Pi_x$ is a Poisson random variable with parameter $x$. So,
$M_k(\infty-)=0$. It is also not hard to see that
$F_k(x,y)$ is continuous in real $x\ge0$ uniformly in real $y\ge0$ (see the Appendix), so that $M_k(x)$ is continuous in $x\ge0$. So, $M_{k+1}(x)$ attains its maximum in $x\ge0$ (equal $a_k$, by (1.5)) and takes all values in the interval $(0,a_k]$. Now Theorem 1 follows by (3). $\qquad\Box$

**Appendix.** Similarly to (2),
\begin{equation*}
\partial_x F_k(x,y)=F_{k+1}(x,y)-F_k(x,y).
\end{equation*}
for real $x,y$. Therefore and because $0\le F_k\le1$, we have $|\partial_x F_k(x,y)|\le1$ for real $x,y$, so that $F_k(x,y)$ is indeed continuous in real $x\ge0$ uniformly in real $y\ge0$.

**Added:** Let us now show that
\begin{equation*}
a_k=c_{k+1},\quad\text{where}\quad c_k:=\frac{k^k}{k!}\,e^{-k}
\sim\frac1{\sqrt{2\pi k}}
\end{equation*}
as $k\to\infty$. To this end, note first that
\begin{equation*}
c_{k+1}/c_k=(1+1/k)^k/e<1,
\end{equation*}
and so, $c_k$ is decreasing in $k$.
So, recalling (4), we have
\begin{equation*}
M_k(x)\le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_{k+j}
\le\sum_{j=0}^\infty \frac{x^j}{j!}e^{-x}\,c_k=c_k=M_k(0).
\end{equation*}
Thus, in view of (1.5) and (2),
\begin{equation*}
a_k=\max_{x\ge0}M_{k+1}(x)=M_{k+1}(0)=c_{k+1},
\end{equation*}
as desired.

In particular, for $k=0,1,2,3$ the values of $a_k$ are $\approx0.367879, 0.270671, 0.224042$.