Does the fibres being equal dimensional imply flatness?

This is true provided that the source is Cohen--Macaulay (e.g. regular) and the base is regular. This is sometimes called miracle flatness. (It can be found as Thm. 23.1 of Matsumura's CRT.)

It generalizes the fact that a non-constant map from a reduced curve to a smooth curve is necessarily flat.

(In the case when the fibre dimension is zero, it is related to the Auslander--Buchsbaum theorem, which says that a f.g. faithful Cohen--Macaulay module over a regular local ring is necessarily free.)

Added: As Georges Elencwajg points out in a comment below, in the context of miracle flatness I should require that we are working with locally finite type schemes over a field.

You are right to be skeptical because the result is false.
For example if $X\subset \mathbb A^2$ is the cusp defined by $y^2=x^3$, its normalization $f:\mathbb A^1=Y\to X: t\mapsto (t^2,t^3)$ is not flat although $f$ is proper and all of its fibers consist of exactly one point and thus have the same dimension, namely $0$ .