Question: (may be silly question but), Does there exist two fiber bundles (or in particular vector bundles) whose total spaces are the same but base spaces are different?

Answer: 
To complement Samuel T's example, and to give an example of you's construction, here's a family of examples which came up a lot in my thesis.

First, a general fact: If $H$ is a compact Lie group acting a manifold $M$ freely (i.e., the only element of $H$ which fixes any point of $M$ is the identity element), then the orbit space $M/H$ has the structure of a smooth manifold in such a way as the projection $M\rightarrow M/H$ is a principal fiber bundle with fiber $H$.

If $M = S^3\times S^3$, and $H = S^1\times S^1$, then there are linear actions of $H$ on $M$ with quotient $S^2\times S^2$, or $\mathbb{C}P^2\sharp - \mathbb{C}P^2$ or $\mathbb{C}P^2\sharp \mathbb{C}P^2$ (and no other quotients are possible), where $-X$ denotes $X$ with reversed orientation. So, we can get three different bases while the total spaces and fibers are the same (and fibers connected, as Samuel T wanted).

If $M = S^3\times S^3\times S^3$ and $H = S^1\times S^1\times S^1$, then Totaro has shown that there are infinitely many different bases of the form $M/H$, even up to rational homotopy equivalence. So this is an example of infinitely many different bases for the same total space, same fiber, and with the fiber connected.

If you insist the the fibers be simply connected, then block embedding of $SU(2)$ into $SU(4)$ and the diagonal block embedding of $SU(2)$ into $SU(4)$ gives rise to 2 different (up to homotopy) bases, giving fiber bundles with total space $SU(4)$ and fibers $SU(2)$ (2 connected), and different bases.
Let $G$ be a topological group, $H$ and $K$ closed subgroups admitting a local cross section (see Steenrod, section 7). Then $H\rightarrow G\rightarrow G/H$ and $K\rightarrow G\rightarrow G/K$ are fibre bundles.

In Martin Wanvik's example, our topological group is $\mathbb{R}^3$, which shares the very nice property that $\mathbb{R}^3/\mathbb{R}^1\cong \mathbb{R}^2$ and vice versa.