How many elements in $GL_2(\mathbb{F}_p)$ are conjugate to $\begin{pmatrix} \lambda & 0 \\ 0 & \mu\end{pmatrix}$ for fixed, distinct $\lambda, \mu \in \mathbb{F}_p$, $p$ prime?
I tried arguing that we have $\frac{p^2-1}{p-1}$ choices for the first eigenvector (division due to rescaling), then $\frac{p^2-p}{p-1}$ for the second eigenvector similarly. This is for any pair of eigenvalues, so divide through by $\frac{1}{(p-1)^2}$ to get this particular pair $\lambda, \mu$. So,
$$\frac{(p^2-1)(p^2-p)}{(p-1)^4}.$$
But this is not an integer for certain $p$ (try $p=7$), so it can't be right.
What am I doing wrong?
It is not clear what you mean in your question by "for any pair of eigenvalues", nor is it clear what you mean in your comment by "for the particular choice of eigenvalues".
In any case, the final division is incorrect. As you have correctly stated, we have $\frac{p^2 - 1}{p-1}$ choices for the eigenvector (up to scaling), corresponding to the specific given eigenvalue $\mu$. Then, we have $\frac{p^2 - p}{p-1}$ choices for the second eigenvector, corresponding to the (distinct) eigenvalue $\lambda$. Once the eigenspaces for $\mu$ and $\lambda$ have been selected, the corresponding transformation is completely and uniquely specified. So, your total should be $$ \frac{(p^2 - 1)(p^2 - p)}{(p-1)^2} = p(p+1). $$
Another justification of the formula, per my comment below.
There are $p+1$ distinct one-dimensional subspaces of $\Bbb F_p$. We select one of these subspaces as our eigenspace associated with $\mu$ and another as our eigenspace associated with $\lambda$, leading to our answer of $(p+1)p$.
To see that there are $p+1$ distinct one dimensional subspaces, it suffices to note that $p$ of these subspaces are spanned by a vector of the form $(1,k)$ for some $k \in \Bbb F_p$ and the remaining subspace is spanned by the vector $(1,0)$.