If X and Y be two topological spaces such that n-th homotopy group of X and (n+1)-th homotopy group of Y are isomorphic for all natural number n. Does it imply that X is homotopy equivalent to the loop space of Y.
Certainly it is not true. If you take a non contractable space whose all homotopy groups are zero and other space to be a point. But I want some interesting example such that homotopy groups are at least not trivial for some n.
The following is one simple example that easily generalises to give you a lot of examples of what you are looking for. It's simple and a nice way to see how homotopy and (co)homology interact.
Let $\mathbb{C}P^\infty\simeq K(\mathbb{Z},2)$ be the complex projective plane and $K(\mathbb{Z},4)$ be the integral Eilenberg-Mac lane space in degree $4$. Let $x\in H^2\mathbb{C}P^\infty$ be a generator and let $E$ be the homotopy fibre in the following sequence
$$E\rightarrow \mathbb{C}P^\infty\xrightarrow{x^2}K(\mathbb{Z},4).$$
Then it is easy to see that $\pi_2E\cong\mathbb{Z}\cong\pi_3E$ and $\pi_kE=0$ for $k\neq 2,3$. Thus $E$ has the same homotopy groups as $\mathbb{C}P^\infty\times K(\mathbb{Z},3)$. However, inspection of the Serre spectral sequence for the previous fibring shows that $H^4E=0$, whilst $H^4(\mathbb{C}P^\infty\times K(\mathbb{Z},3))\cong\mathbb{Z}\{x^2\times 1\}$. Hence $E\not\simeq \mathbb{C}P^\infty\times K(\mathbb{Z},3)$.
Of interest is that fact that it does hold that
$$\Omega E\simeq \Omega(\mathbb{C}P^\infty\times K(\mathbb{Z},3))\simeq S^1\times K(\mathbb{Z},2)$$
since looping kills the decomposable $x^2$.
If you want a really nice example of what you are looking for, then I suggest the paper Loop Structures On the Homotopy Type of $S^3$ by David Rector. $S^3\cong Sp_1$ is a compact Lie group which admits a delooping $B$ satisfying $S^3\simeq \Omega B$. Rector shows that there are infinitely many such space $B$, and goes on to list a complete set of invariants that classify them, at least in a $p$-local setting.