Find the PDF of $Y = X^2 + 3$ where $X \sim Poisson(\lambda)$. I am following the CDF method to calculate the PDF oF Y. Up to this, I have done:
$$F_Y = P(Y \leq x)$$
$$F_Y = P(X^2+3 \leq x)$$
$$F_Y = P(X \leq \sqrt{x-3)}$$
$$F_Y = e^{-\lambda}.\sum_{k=0}^{\sqrt{x-3}} \frac{\lambda^k}{k!}$$
Now I have to differentiate the CDF to get PDF.
$$f_Y  = \frac{\partial F_Y}{\partial x}$$
Now how can I proceed from here? In the continuous case, I can use Leibniz rule but I'm not sure about the discrete case.

The CDF methods you showed is for continuous distribution. With discrete laws the pdf (more precisely the pmf, probability mass function) is not the derivative of the CDF.
With a discrete distribution, it is enough to directly calculate the pmf:
$$\mathbb{P}[Y=y]=\mathbb{P}[X^2+3=y]=\mathbb{P}[X=\sqrt{y-3}]=\frac{e^{-\lambda}\lambda^{ \sqrt{y-3}}}{ (\sqrt{y-3})!}\mathbb{1}_{\{3;4;7;12,...\}}(y)$$
