I am trying to show that $$\mathcal{L}\left(\sum_{n=0}^{\infty} f(nT)\delta(t-nT)\right)=\sum_{n=0}^{\infty} f(n)z^{-n},$$ where $n\in\mathbb{Z^+}$, $z\in\mathbb{C}$ and $\delta$ is the Dirac delta function. Note tht $f(nT)$ and $f(n)$ do not denote the same function.
So far, I have shown that \begin{align} \mathcal{L}\left(\sum_{n=0}^{\infty} f(nT)\delta(t-nT)\right)&=\int_{0}^{\infty}e^{-st}\left(\sum_{n=0}^{\infty} f(nT)\delta(t-nT)\right) dt \\ &=\sum_{n=0}^{\infty} f(nT)\int_{0}^{\infty}e^{-st}\delta(t-nT) \ dt \\ &=\sum_{n=0}^{\infty}f(nT)\mathcal{L}\left(\delta(t-nT)\right) \ dt \\ &=\sum_{n=0}^{\infty}f(nT)e^{-snT} \tag{1}. \end{align} A hint on further progression would be appreciated. I do not require a full result.
Update: Can we assume that $f(n)=f(nT)$ as $T$ denotes the period?
The steps you have taken are correct. As you have stated, the function $f$ on the right hand side (as $f(n)$) is not the same function $f$ on the left hand side (as $f(nT)$). To be slightly more clear, let us define $f[n] := f(nT)$ (with square brackets) and $z := e^{sT}$ then \begin{align} \sum_{n=0}^{\infty} f(nT) e^{-snT} &= \sum_{n=0}^{\infty} f[n] (e^{sT})^{-n} \\ &= \sum_{n=0}^{\infty} f[n] z^{-n}. \\ \end{align}
This identity is mainly used in sampling continuous time signals with $T$ being the sampling period (this may be different from the period of the $f$, if $f$ is periodic). The $f[.]$ is referred to as the discrete-time sample. For further information, you may refer to the following question from Signal Processing Stack Exchange: sampling with Laplace Transform.