Intervals of convolution product
An easy way to deduce the nonzero range of discrete-time convolutions is the following:
Consider convolution of $x[n]$ with $y[n]$: $$z[n] = x[n] \star y[n] .$$
Let the shorther sequence $x[n]$ has a nonzero range of $M_1 \leq n \leq M_2$ , while the nonzero range of longer $y[n]$ be $N_1 \leq n \leq N_2$ .
1- Expand $x[n]$ as weighted impulses: $x[n] = \sum_{k=M_1}^{M_2} x[k] \delta[n-k]$ .
2- Distribute the convolution over expanded $x[n]$:
$$ (\delta[n-M_1]+...+\delta[n-M_2]) \star y[n] = \delta[n-M_1] \star y[n] + ... + \delta[n-M_2] \star y[n] $$
3- Consider the first impulse $\delta[n-M_1]$, and the last impulse $\delta[n-M_2]$.
4- Apply shifting property for those two impulses :
$$ \delta[n-M_1] \star y[n] = y[n-M_1]  ~~~,~~~  \delta[n-M_1] \star y[n] = y[n-M_2] .$$
5- Determine the nonzero ranges of $y[n-M_1]$ and $y[n-M_2]$:
$y[n]$ is nonzero over $N_1 \leq n \leq N_2$, then $y[n-M_1]$ will be nonzero for $$ N_1 \leq n-M_1 \leq N_2  ~~~~~~\Rightarrow ~~~~~~  N_1 + M_1 \leq n \leq N_2 + M_1 .$$
Similarly for $y[n-M_2]$ the nonzero range will be: $$ N_1 \leq n-M_2 \leq N_2  ~~~~~~\Rightarrow ~~~~~~ N_1 + M_2 \leq n \leq N_2 + M_2 .$$
One can see that the output $z[n]$ has a nonzero range that begins at the leftmost sample of $y[n-M_1]$, and ends at the rightmost sample of $y[n-M_2]$. Yielding:
$$N_1+M_1 \leq n \leq N_2 + M_2 .$$
Length of $z[n]$ can be found as: $$L_z = L_x + L_y -1 = N_2+M_2 - N_1-M_1 + 1 .$$
Applying this to yourcase, $N_1 = 0$, $M_1 = 0$, $N_2 = 3$ and $M_2=2$ which yields the nonzero range to be
$$ 0 + 0 = 0 \leq n \leq 5 = 2 + 3 .$$