Yes.

The space of rational numbers $X=\mathbb{Q}$ is an instance.

We can view $X$ as a countable union of countably many disjoint copies of $\mathbb{Q}$.

Any nonempty subset $A$ of those copies (that is, taking all or none of each
copy) is a retract of $X$, since we can map the unused copies to a
fixed copy, and this is continuous. And there are $2^{\aleph_0}$ many such $A$, so we've got enough retracts.

A similar idea works with copies of other spaces, and one can make uncountable examples this way.

More examples:

Every infinite ordinal $\lambda$, under the order topology. Any closed subset $A\subset\lambda$ is a retract, since you can map each $\alpha<\lambda$ to the next element of $A$, that is, the smallest $\beta\in A$ with $\alpha\leq\beta$. This is continuous. And there are $2^{|\lambda|}$ many closed subsets of $\lambda$.

The long rational line, $\left([0,1)\cap\mathbb{Q}\right)\cdot\omega_1$. This has size $\omega_1$, and yet every closed subset of $\omega_1$ gives rise to a retract (by taking the point $0$ in that interval). So there are $2^{\omega_1}$ many retracts.