As pointed out in my comment, an example would be given by the counting measures on $\mathbb Q \cap [0,1]$ and $(\sqrt{2} + \mathbb Q)\cap [0,1]$, respectively, on the compact metric space $X = [0,1]$.
Note that the same idea actually works for any compact metric space $X$ which has no isolated points.
Since you also asked about an example where the measure spaces are finite:
You can simply take "weighted measures", i.e. if $\{q_n\}_{n \in \mathbb N}$ is a enumeration of $\mathbb Q$, define a function
$$f(x) = \begin{cases} 2^{-n} & \text{if } x = q_n \\ 0 & \text{otherwise} \end{cases} $$
Now the weighted measure is given by $d\tilde \mu = f \, d\mu$, where $\mu$ is the counting measure on the rationals. This will then be finite
$$\int_\mathbb{R} \; d\tilde\mu = \int_\mathbb{R} f \; d\mu = \sum_{n = 1}^\infty 2^{-n} = 1$$
Let $\delta_n:=\frac{2n}{(2n+1)n(n-1)}$, and $g_n$ defined by
$$g_n(x)=\begin{cases}
n&\mbox{ on }\left(\frac kn-\frac{\delta_n}2,\frac kn+\frac{\delta_n}2\right), 1\leq k\leq n-1\\\
\mbox{ linear }&\mbox{ on }\left(\frac kn-\frac{\delta_n}2-\frac{\delta_n}{2n},\frac kn-\frac{\delta_n}2\right)\\\
\mbox{ linear }&\mbox{ on }\left(\frac kn+\frac{\delta_n}2,\frac kn+\frac{\delta_n}2+\frac{\delta_n}{2n}\right)\\\
0&\mbox{ elsewhere}.
\end{cases}$$
We have that the measure of the support of $g_n$ is $(n-1)(1+1/n)\delta_n$ which converges to $0$, and $\int_{[0,1]}g_ndm=1$, except miscomputation. We can write
$$\left|\int_{[0,1]}g_nfdm-\frac 1n\sum_{k=1}^nf\left(\frac kn\right)\right|\leq 2\delta_n \lVert f\rVert_{\infty}+\operatorname{mod}(f,\delta_n)\frac{n^2}{\delta_n},$$
where $\operatorname{mod}(f,\delta):=\sup\{|f(x)-f(y)|,x,y\in I, |x-y|\leq \delta \}$.
So the three conditions are full-filled by $m$ and $m$ are of course not singular.
Best Answer
Welcome to MSE!
Hint:
The Ergodic Theorem tells you that for any $f : L^1(X)$ and for almost every $x$
$$\lim_{n \to \infty} \frac{1}{n} \sum_{i<n} f(T^ix)$$
equals something. Notice this limit doesn't depend on $\mu$ or $\nu$, whereas the something I'm alluding to does.
Can you leverage this (say by choosing $f = \chi_A$ to be a characteristic function) in order to prove $\mu = \nu$?
I hope this helps ^_^