What is $\dim_K K[X,Y]/\frak{m}^n$, where $\frak{m}$ $= (X-a,Y-b)$?

I am stuck at the following exercise:

Let $K$ be an (algebraically closed) field. What is $\dim_K K[X,Y]/\mathfrak{m}^n$, where $\mathfrak{m}= (X-a,Y-b)$?

This exercise is very similar to the one solved here, but since we here we are working with $(X-a,Y-b)$ instead of $(X,Y)$ we need to make some modifications: My idea would be to substite

$$X^{\prime} := X-a \quad \text{ and } \quad Y^{\prime} := Y-b$$

and then use $\dim_k(k[X,Y]/I^n)=\frac{n(n+1)}{2}$, which was established in the linked question. However, I am not sure if using such a substitution is possible (or how to argue it). Could you please tell me if this works?

Remark: I understand that $f: K \rightarrow K, x \mapsto x-a$ is a bijection.

There is a unique unital $k$-algebra isomorphism $$k[X,Y]\to k[X,Y]: X \mapsto X-a, Y \mapsto Y-b.$$

This isomorphism sends one ideal to the other, so it induces an isomorphism between the corresponding quotient algebras.

Then, simply take dimensions.