For every sufficiently large $m$ there exists $k$ such that $m = k + \tau(k)$ Let $\tau(k)$ , be the number of positive divisors of natural number $k$. Is it true, that there exists $n_0$ , such that for every $m\geq n_0$ there exists $k  \in\mathbb{N}$ such that:
$$
m = k + \tau(k)
$$

I have tried to use the following formula for $\tau$:
$$
\tau(p_1^{k_1}\ldots p_{s}^{k_s}) = (k_1 +1)\cdot\ldots \cdot(k_s +1),
$$
Where $p_1, \ldots, p_s$ are different prime numbers.

Intuitively, I think that the answer will be no. So, we can assume the contrary (that such $n_0$ exists) and try some infinite series of numbers (primes, factorials, primorials, etc.), which can't be written in the form $k + \tau(k)$ for every $k$. But my attempts weren't successful.

So, I will be grateful for hints and ideas.
contest problem related to divisor function 
Look at Loppukilpailija's answer. I think he solved your problem negatively.
We are looking at, a specific composition. Included, is a multiplicative partition( number of divisors, in this case). Highly composite numbers are likely to have the highest number of multiplicative partitions due to highest divisor number up to a point. These may not be so dense on the naturals however. 12 divisors comes about in one of 4 ways:$$p^{11}\\p^5q\\p^3q^2\\p^2qr$$