Extend your diagram using $\newcommand{\Ex}{\mathrm{Ex}}\newcommand{\Sd}{\mathrm{Sd}}\Ex^\infty$:

$$\begin{CD} \partial\Delta[n] @>>> X @>{\sim}>> \Ex^\infty X \\ @VVV @VV{f}V @VV{\Ex^\infty f}V \\ \Delta[n] @>>> Y @>>{\sim}> \Ex^\infty Y \textrm{.} \\ \end{CD}$$

Then $f$ is a weak equivalence if and only if $\Ex^\infty f$ is if and only if there is a relative lift in the outer rectangle (since $\Ex^\infty X$ and $\Ex^\infty Y$ are Kan complexes). Since such a lift involves only finitely many simplices of $\Ex^\infty X$ and $\Ex^\infty Y$, there is a finite $k$ such that it actually lives in

$$\begin{CD} \partial\Delta[n] @>>> X @>{\sim}>> \Ex^k X \\ @VVV @VV{f}V @VV{\Ex^k f}V \\ \Delta[n] @>>> Y @>>{\sim}> \Ex^k Y \textrm{.} \\ \end{CD}$$

Since $\Ex^k$ is adjoint to $\Sd^k$, such a diagram translates to

$$\begin{CD} \Sd^k\partial\Delta[n] @>>> \partial\Delta[n] @>>> X \\ @VVV @VVV @VV{f}V \\ \Sd^k\Delta[n] @>>> \Delta[n] @>>> Y \textrm{.} \\ \end{CD}$$

So in the end $f$ is a weak equivalence if and only if there is a $k$ such that the latter diagram admits a "relative lift". However, you have to be careful here since a homotopy $\Delta[n] \times \Delta[1] \to \Ex^k Y$ translates to a map $\Sd^k(\Delta[n] \times \Delta[1]) \to Y$ which is different than $\Sd^k\Delta[n] \times \Delta[1] \to Y$ or $\Sd^k\Delta[n] \times \Sd^k\Delta[1] \to Y$ so "relative lift" has a slightly different meaning now.

$\newcommand{\Ex}{\mathrm{Ex}}\newcommand{\Sd}{\mathrm{Sd}}$Consider the square

$$\begin{CD} X @>{\sim}>> \Ex^\infty X \\ @V{f}VV @VV{\Ex^\infty f}V \\ Y @>>{\sim}> \Ex^\infty Y \textrm{.} \\ \end{CD}$$

The map $f$ is a weak equivalence if and only if $\Ex^\infty f$ is if and only if there is a relative lift in every diagram of the form

$$\begin{CD} \partial\Delta[n] @>>> \Ex^\infty X \\ @VVV @VV{\Ex^\infty f}V \\ \Delta[n] @>>> \Ex^\infty Y \textrm{.} \\ \end{CD}$$

Since such a lifting problem and its solution involve only finitely many simplices of $\Ex^\infty X$ and $\Ex^\infty Y$, there is a $j$ such that the lifting problem actually lives in

$$\begin{CD} \partial\Delta[n] @>>> \Ex^j X \\ @VVV @VV{\Ex^j f}V \\ \Delta[n] @>>> \Ex^j Y \\ \end{CD}$$

and a $k$ such that its solution lives in

$$\begin{CD} \partial\Delta[n] @>>> \Ex^j X @>>> \Ex^{j+k} X\\ @VVV @VV{\Ex^j f}V @VV{\Ex^{j+k} f}V \\ \Delta[n] @>>> \Ex^j Y @>>> \Ex^{j+k} Y \textrm{.} \\ \end{CD}$$

By adjointness, this means that $f$ is a weak equivalence if and only if for every $j$ and every lifting problem

$$\begin{CD} \Sd^j\partial\Delta[n] @>>> X \\ @VVV @VV{f}V \\ \Sd^j\Delta[n] @>>> Y \\ \end{CD}$$

there is a $k$ and a "relative lift" in

$$\begin{CD} \Sd^{j+k}\partial\Delta[n] @>>> \Sd^j\partial\Delta[n] @>>> X \\ @VVV @VVV @VV{f}V \\ \Sd^{j+k}\Delta[n] @>>> \Sd^j\Delta[n] @>>> Y \textrm{.} \\ \end{CD}$$

However, you have to be careful here since a homotopy $\Delta[n] \times \Delta[1] \to \Ex^{j+k} Y$ translates to a map $\Sd^{j+k}(\Delta[n] \times \Delta[1]) \to Y$ which is different than $\Sd^{j+k}\Delta[n] \times \Delta[1] \to Y$ or $\Sd^{j+k}\Delta[n] \times \Sd^{j+k}\Delta[1] \to Y$ so "relative lift" has a slightly different meaning now.