Such an $f$ only exists in two cases: when $k=1$ (when $\beta Sq^1 = 0$) and when $k=0$ (when $\beta Sq^0 = \beta$). Here is a proof, which will take a little work with the Steenrod algebra.

Let's suppose that $X$ and $Y$ are either $H\Bbb Z$ or $H\Bbb Z/2$ (not necessarily the same). Then the map
$$
[X,Y]_t \to Hom_{A^*}(H^*Y, \Sigma^t H^*X),
$$
sending a map to its (graded) effect on cohomology, is an isomorphism for $t \neq 0$. If one of $X$ or $Y$ is $H\Bbb Z/2$, it is an isomorphism for all $t$.

Let's prove this. This map is the edge morphism in the Adams spectral sequence
$$
Ext_{A^*}^s(H^*Y, \Sigma^t H^*X) \Rightarrow ([X,Y]^\wedge_2)_{t-s}
$$
and so it suffices to show that the Adams spectral sequence degenerates.

If $Y = H\Bbb Z/2$, then $H^* Y$ is free over $A^*$ and there are no higher Ext-groups.

If $Y = H\Bbb Z$, then $H^* Y = A^* / Sq^1$, which has a free resolution
$$
\cdots \to A^* \xrightarrow{\cdot Sq^1}A^* \xrightarrow{\cdot Sq^1} A^* \to A^*/Sq^1 \to 0
$$
(using the basis of $A^*$ in terms of admissible monomials). Applying $Hom_{A^*}(-, \Sigma^* H^*(X))$ to this resolution, we get the chain complex
$$
0 \to \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \xrightarrow{Sq^1 \cdot} \Sigma^* H^*(X) \to \cdots
$$
which gives us the Ext-groups (in particular, we could interpret this as the Bockstein spectral sequence from mod-2 cohomology to integral cohomology). If $X = H\Bbb Z/2$, then the basis of admissible monomials shows that this is exact except in degree zero, where it recovers $ker(Sq^1 \cdot -) = Im(Sq^1 \cdot -)$. If $X = H\Bbb Z$, then $H^*(X)$ has a basis consisting of admissible monomials we similarly find that this is exact *except* in degree zero: $A^* / Sq^1$ is a direct sum of $\Bbb Z/2$ in degree zero with a bunch of free modules over $\Bbb Z/2[Sq^1] / (Sq^1 \cdot Sq^1)$ on the admissible monomials $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$. Therefore, the higher Ext-groups vanish except for those contributing to $[X,Y]_0 = [H\Bbb Z, H\Bbb Z]_0$. The degree-zero part $Hom(A^*/Sq^1, A^*/Sq^1) = ker(Sq^1 \cdot -)$ has a basis consisting of $1$ and admissible monomials of the form $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$.

In particular, we found in the course of the proof that the map
$$
[H\Bbb Z, H\Bbb Z]_n \to [H\Bbb Z, H\Bbb Z/2]_n,
$$
sending $f$ to $\rho_2 \circ f$, is injective for $n \neq 0$. The image is parametrized by those elements of $A^* / Sq^1$ which are sums of admissible monomials $Sq^{2k+1} Sq^{n_1} \cdots Sq^{n_r}$; this corresponds to the cohomology operation $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$.

Thus:

The set $([H\Bbb Z, H\Bbb Z]_n)^\wedge_2$ is $2$-torsion for $n > 0$, and has a basis consisting of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ admissible monomials in the Steenrod algebra with $n_r > 1$.

The set $[H\Bbb Z/2, H\Bbb Z]_n$ has a basis of elements of the cohomology operations $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r}$ over $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ admissible monomials in the Steenrod algebra.

The subset of elements of the form $f \beta$ are thus sums of elements of the form
$$
\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} \rho_2 \beta=
\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1.
$$
The image of right multiplication by $\beta$, therefore, has a basis consisting of sums of elements of the form $\beta Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$, where $Sq^{2k} Sq^{n_1} \cdots Sq^{n_r} Sq^1$ is an admissible monomial in the Steenrod algebra. In particular, $\beta Sq^n$ is not of this form unless $n=1$ (when it is zero) or $n=0$ (when it is $\beta$).