Suppose that $f$ cannot be written as $f(x_1,x_2,\dots,x_n)=h(g(x_1,x_2,\dots,x_n))$ with $h\in\mathbb C[x]$ of degree $\ge2$ and $g\in\mathbb C[x_1,x_2,\dots,x_n]$. For $m\in\mathbb C$ let $a_m$ be the number of irreducible factors of $f(x_1,x_2,\dots,x_n)+m$. Then Ewa Cygan proved (see here for a review and reference) that $\sum_{m\in\mathbb C}(a_m-1)\le\deg f-1$. (The case $n=2$ was shown previously by Josef Stein, see here.)

So, as hinted in the comments, there can be infinitely many $m$ with $f(x_1,x_2,\dots,x_n)+m$ reducible. But in that case $f$ has a rather specific shape.

The question also asks to decide if there is at least one $m$ making $f+m$ reducible. Here is a sketch on how to find the $m$'s in case that there are only finitely many: Suppose that $f+m=uv$ with non-constant polynomials. Pick $\bar x\in\mathbb C^n$ with $u(\bar x)=v(\bar x)=0$. (There probably is no such $\bar x$, so one should better work in the projective completion.) Then the partial derivatives $\frac{\partial f}{\partial x_i}=\frac{\partial u}{\partial x_i}v+u\frac{\partial v}{\partial x_i}$ vanish in $\bar x$.

So the recipe is as follows: Determine all (usually finitely many) common roots $\bar x$ of the partial derivatives $\frac{\partial f}{\partial x_i}$, set $m=-f(\bar x)$, and check if $f(x_1,\dots,x_n)+m$ is reducible.

As said above, one should better work projectively in order to avoid missing solutions $m$ coming from singularities at infinity.