$X_1,\dots, X_n$ is a simple random sample of $X\sim \operatorname{Poisson}(\lambda)$. What is the optimal critical region $\alpha$ for $H_0:\lambda =\lambda_0$ against $H_1:\lambda=\lambda_1$?
I would use the Neyman-Pearson theorem to solve it. $$ \frac{\lambda_0}{\lambda_1} = \frac{(\lambda_0^{\sum x_i} e^{-\lambda_0}) / (\prod_{i=1}^n x_i!)}{(\lambda_1^{\sum x_i}e^{-\lambda_1})/(\prod_{i=1}^n x_i!)} = \frac{\lambda_0^{\sum x_i}e^{\lambda_1}}{\lambda_1^{\sum x_i}e^{\lambda_0}}\leq k$$
Taking logarithms
$$ \ln \left( \lambda_0^{\sum x_i}e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}e^{\lambda_0} \right)\leq \ln(k)\\ \ln \left( \lambda_0^{\sum x_i}\right)+\ln\left(e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}\right)-\left(e^{\lambda_0} \right)\leq \ln(k)\\ \left(\sum x_i\right)\ln (\lambda_0)+\lambda_1 -\left( \sum x_i \right)\ln \left( \lambda_1\right)-\lambda_0\leq \ln(k)\\ \left(\sum x_i\right)\left(\ln (\lambda_0) -\ln \left( \lambda_1\right)\right)\leq \ln(k)+(\lambda_0-\lambda_1)\\ \sum x_i\leq \frac{\ln(k)+(\lambda_0-\lambda_1)}{\ln (\lambda_0) -\ln \left( \lambda_1\right)} $$
Now what would be the optimal critical region? This will depend of $\lambda_0,\lambda_1$ according to some criterion I cannot determine from the last equation.
You have made use of an unstated assumption that $\lambda_0>\lambda_1.$
You concluded that the rejection criterion is that $\sum_{i=1}^n X_i < c$ where $c$ is the critical value.
If the level of the test is $\alpha$ then the critical value $c$ must satisfy \begin{align} & \Pr\left( \sum_{i=1}^n X_i \le c \,\middle|\, \lambda = \lambda_0 \right) \le \alpha \\ \text{and} & \text{ (since only integer values are involved) } \\ & \Pr\left( \sum_{i=1}^n X_i \ge c+1 \,\middle|\, \lambda=\lambda_0 \right) >\alpha. \end{align} Given $\lambda_0$ and $\alpha,$ and recalling that $X_1+\cdots+X_n\sim\operatorname{Poisson}(n\lambda_0)$, use those two inequalities to find $c.$