If you are liberal enough as to what you consider to be "a bundle", then this is true. The reason is that the symmetric product $$ G := SP^\infty(S^{n-1}) \simeq K(\mathbb{Z},n-1)$$ is a topological abelian group, and principal $G$-bundles are classified by maps to $BG \simeq K(\mathbb{Z}, n)$. Thus if $f : X \to Y$ induces a bijection between all fibre bundles, then it is a (co)homology equivalence.
Considering all principal $H$-bundles for $H$ a discrete group, one also sees that $f$ induces an isomorphism on $\pi_1$ (as $f_* : \mathrm{Hom}(\pi_1(X), H) \cong \mathrm{Hom}(\pi_1(Y), H)$ for all groups $H$ means $\pi_1(X) \to \pi_1(Y)$ is an isomorphism).
Thus $f$ has to be a weak homotopy equivalence. As the question as asked for compact smooth manifolds, which are homotopy equivalent to CW-complexes, $f$ is a homotopy equivalence.
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As TG points out in the comments, $\pi_1$ equivalence and homology equivalence are not enough to deduce weak homotopy equivalence without a simpleness assumption.
Thus: for each discrete group $H$ and $H$-module $V$ we can form $$G := K(V, n-1) \rtimes H$$ a topological group. A concrete model for $K(V, n-1)$ can be taken to be $SP^\infty_V(S^{n-1})$, the symmetric product of the sphere with labels in $V$, on which $H$ acts on the labels. A principal $G$-bundle over $X$ is classified by a map to the total space of the fibration $$K(V, n) \to BG \to BH,$$ and this consists of a homomorphism $f:\pi_1(X) \to H$ and a twisted cohomology class in $H^n(X; f^*V)$.
If I am not mistaken again, a map inducing a $\pi_1$-iso and an iso on cohomology with all local coefficients is indeed a weak equivalence.