With regards to "poles and zeros", it's not a time domain input that leads to a zero or infinity, but the transfer function itself as the Laplace Transform of a time domain signal. That time domain signal is typically the impulse response of a system or filter. The Laplace Transform provides us a frequency domain view; in this case the "frequency domain" consists of all the spinning phasors $e^{-j\omega t}$ just as we have with the Fourier Transform as well as the greater family of those phasors that grow and decay with time: $e^{-\sigma t} e^{-j\omega t} = e^{-st}$.
Just as the Fourier Transform decomposes an arbitrary function into it's $e^{j\omega t}$ components as spinning phasors with constant magnitude (the real and imaginary components being the sinusoids we typically associate with "frequency"), the poles in the Laplace Transform correspond to components in the time domain waveform (the impulse response of the system) as phasors that grow and decay with time: $e^{s t}$. The zeros correspond to any functions that would be orthogonal to the impulse response.
For example: if the transfer function was $H(s)= \frac{1}{s+2}$, the impulse response as the inverse Laplace Transform of $H(s)$ is $h(t) = e^{-2t}u(t)$. ($u(t)$ is the unit step function, for the unilateral Laplace Transform with the assumption that all functions we are using are causal). Here we see that the pole at $s=-2$ indicates the correlation of the function $e^{st} = e^{-2t}$ with $h(t)$, in this case it is completely described by that one single component; we have decomposed $h(t)$ into components of $e^{st}$ which in this case was trivial given $h(t)$. But this is to demonstrate how that pole at $s=-2$ indicates there is a component of $e^{-2t}$ in $H(s)$. (The complete inverse Laplace Transform of $H(s)$ as given by the OP with one pole and one zero is $h(t) = \delta(t)+e^{-2t}u(t)$).
Thus the Transfer Function, as the Laplace Transform of the systems time domain impulse response, provides us another representation of that same function, mapped to the frequency domain (and the greater frequency domain that not only has phase changing with time, but also has magnitude changing with time). We can use this to determine outputs for various inputs, and also to assess characteristics about the system such as stability, settling time, rate of change, etc.
The Laplace Transform of the output for any transfer function would be the product of the Laplace Transform of the input with the Transfer Function.
$$Y(s) = H(s)X(s)$$
The time domain output would then be the inverse Laplace Transform of $Y(s)$.
If we want to determine an input that would cause the output to be $0$ or $\infty$, first we must clarify that the time domain result is changing with time; it will either grow to infinity, or shrink to zero, or pass through zeros as the waveform proceeds with time. What the OP may be interested in is inputs that would result in $0$ or $\infty$ as the "Final Value", or the value we expect to have at the output at $t=\infty$.
For this we can use the Final Value Theorem:
$$y(t=\infty) = \lim_{s\rightarrow0}(sY(s))$$
For example with the OP's case:
If the input $x(t)$ is an impulse, the Laplace Trasform is $X(s)=1$ and the output $Y(s)=X(s)H(s) = H(s)$ is the Laplace Transform of the "impulse response" consistent with what I introduced about the trasfer function. The final value of this impulse response, at $t=\infty$ would be:
$$y(t=\infty) = \lim_{s\rightarrow0}\bigg(s\frac{s+3}{s+2}\bigg) = 0\frac{3}{2} = 0$$
Thus an impulse (and infinitely many other inputs) will result in the output being 0, eventually, at $t=\infty$.
An example input that would result in the time domain output going infinity is:
$$X(s) = \frac{s+2}{s^2}$$
As this would result in an output:
$$Y(s) = X(s)H(s) = \frac{s+2}{s^2}\frac{s+3}{s+2} = \frac{s+3}{s^2}$$
And the time domain output at $t=\infty$ would be:
$$y(t=\infty) = \lim_{s\rightarrow0}\bigg(s\frac{s+3}{s^2}\bigg) = \infty$$
The inverse Laplace transform of $X(s)$ in this case is $x(t)=2t+1$, and the resulting output is $y(t) = 3t + 1$.
