Are safe primes $p=2^k \pm s$ with $s$ small less recommandable than others as a discrete log modulus? I take the definition of safe prime as: a prime $p$ is safe when $(p-1)/2$ is prime.

Safe primes of appropriate size are the standard choice for the modulus of cryptosystems related to the discrete logarithm problem, such as Diffie-Hellman.

A modulus $p=2^k \pm s$ with $s$ small makes exponentiation $\bmod p$ appreciably simpler (the problem of quotient estimation in classical Euclidean division all but vanishes, and the cost of modular reductions becomes negligible compared to multiplications). This, and the desire to use a non-arbitrary, compactly stored modulus, makes it tempting to prefer such choice of $p$. Like "the modulus shall be the biggest 2048-bit safe prime, which is $p=2^{2048}-1942289$".

Are there any major reason not to do that, like a much faster algorithm for the discrete logarithm problem?
Short answer: Yes.

The discrete logarithm can be attacked in a multitude of ways: Baby-step giant-step (BSGS), Pollard's Rho, Pohlig-Hellman, and the several variants of Index Calculus, the best of which currently is the Number Field Sieve. 

Let $n$ be the order of the generator of our field $\mathbb{F}_p$; it is $n = p-1$. We are trying to find $x$ given $a$ and $b=a^x$ in the above field.

Pollard's Rho and BSGS

In Baby-step giant step, we are trying to find $i$ and $j$ such that $b(a^{-m})^i = a^j$, where $m = \lceil \sqrt{n} \rceil$. Once we find such a pair, the discrete logarithm $x = i m + j$ follows, as $ b(a^{-m})^i = a^j \Leftrightarrow a^{x - m i} = a^j \Leftrightarrow x \equiv mi + j \pmod{n}$.

To do so, we first compute a table of all $a^j$ for all $j$ up to $m-1$. Then we iterate through all $i$ up to $m-1$, and compare $b(a^{-m})^i$ with $a^j$. Ignoring arithmetic costs, the runtime of this method is at most $2(m-1) = O(\sqrt{n})$ (with the same space requirements).

Due to the large space requirements, BSGS is rarely used in practice. Instead, we turn to Pollard's Rho. The crux of this method is to find a colliding nontrivial pair $(i,j)$ and $(k,l)$ such that $a^ib^j \equiv a^kb^l$. It follows that $x = \frac{k-i}{l-j} \pmod{n}$, since $a^i a^{xj} = a^k a^{xl} \Leftrightarrow a^{i + xj} = a^{k + xl} \Leftrightarrow i + xj \equiv k + xl \pmod{n}$.

So Rho comes down to finding a collision quickly. This can be done with various algorithms, Floyd's being the oldest and best known. The good news is that we can try and find a collision without an enormous table; the not so good news is that the algorithm is probabilistic, although the birthday paradox tells us we should expect a collision in about $\sqrt{n}$ steps.

In any case, these attacks are no good against a safe prime, where the order large enough that $\sqrt{n}$ is computationally unfeasible.

Pohlig-Hellman

The Pohlig-Hellman approach relies on the observation that there is an homomorphism $\phi$ from $a$ and $b$ from their group of order $n$, to the subgroup of order $p_i^{e_i}$ dividing $n$. In general, given $n = p_1^{e_1}p_2^{e_2}\ldots p_m^{e_m}$,

$$
\phi_{p_i^{e_i}}(a) = a^{n/p_i^{e_i}}
$$

This allows us to compute the discrete logarithm of $\phi_{p_i^{e_i}}(a)$ and $\phi_{p_i^{e_i}}(b)$, which really is the discrete logarithm of $a$ and $b$ modulo $p_i^{e_i}$. From this observation, it is a matter of computing the logarithm modulo all prime divisors of $n$ (using the methods in the previous section) and combining them together using the Chinese remainder theorem.

If $n$ has many small prime divisors, i.e., it is smooth, this method is very much faster than Rho or BSGS. In a safe prime, however, this is not the case, since the order $n$ is the product $2q$, for a very large $q$. Pohlig-Hellman doesn't help much here.

Index Calculus

Index Calculus is the basis for the best-performing algorithms to compute discrete logarithms modulo safe primes. Suppose we know the logarithms of $2$ and $3$; finding the logarithm of $12$ is easy: $\log_a12 = 2 log_a2 + log_a3$, since $12$ factors into $2^2\times 3$.

We can generalize this method to arbitrary elements. Start by defining the factor base, i.e., all the primes up to some bound $B$. Then, find the logarithms of all the elements of the factor base (this is the tricky part). Finally, factor $b$ into the factor base, and simply add all the logarithms corresponding to the factorization you find. If $b$ does not factor completely into the factor base, multiply $b$ by some known exponent of $a$ and try again.

Finding the logarithms of all the primes up to $B$ requires some trickery. It has two major steps:


For $k_i \in [1..n]$, find (usually by sieving) at least $\pi(B)$ elements $a^k$ that factor completely into the factor base. Store both $a^{k_i}$ and its complete factorization.
Now we have the linear system (modulo $n$):
$$
\begin{eqnarray}
e_{1,1} \log_a 2 &amp;+&amp; e_{1,2} \log_a 3 &amp;+&amp; \ldots &amp;+&amp; e_{1,{\pi(B)}} &amp;=&amp; {k_1} \\
e_{2,1} \log_a 2 &amp;+&amp; e_{2,2} \log_a 3 &amp;+&amp; \ldots &amp;+&amp; e_{2,{\pi(B)}} &amp;=&amp; {k_2} \\
&amp;&amp;&amp;&amp;\ldots&amp;&amp;&amp;&amp; \\
e_{{\pi(B)},1} \log_a 2 &amp;+&amp; e_{{\pi(B)},2} \log_a 3 &amp;+&amp; \ldots &amp;+&amp; e_{{\pi(B)},{\pi(B)}} &amp;=&amp; {k_{\pi(B)}} \\
\end{eqnarray}
$$
Solving the above linear system gives us the needed logarithms for the factor base.


The runtime of this method, for appropriate choice of $B$, is $\exp{((2+o(1)((\log n)^{1/2}(\log \log n)^{1/2}))}$. This is not strictly polynomial, but is a big improvement on the previous methods.

Number field sieve

The number field sieve is currently the best algorithm for both integer factorization and discrete logarithms over finite fields. For the discrete logarithm, it is analogous to the above index calculus, with a few major modifications:


We are working in the number fields $\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\beta]$ instead of the integers; there is, however, a map from such fields to the integers under some conditions. The number fields are defined by the polynomials $f_1$ and $f_2$ of degree $d_1$ and $d_2$; there must exist an integer $m$ such that $f_1(m) = f_2(m) = 0 \pmod{p}$.
The factor base is formed by the primes in both $\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\beta]$, up to bounds $B_1$ and $B_2$.
During sieving, we look for pairs $(x,y)$ such that $N_{f_1}(x + \alpha y)$ and $N_{f_2}(x + \beta y)$ are $B_1$ and $B_2$-smooth, respectively, where $N_{f_i}$ is given by
$$
N_{f_i}(x + \alpha y) = y^{d_i} f_i(x/y)
$$ 


The speed of the number field sieve hinges on the speed of finding $(x,y)$ with smooth norms $N_{f_i}(x,y)$. In turn, the probability of $N_{f_i}(x,y)$ being smooth is linked to its size: the smaller it is, the more likely it is to be smooth. And in turn, the size of $N_{f_i}(x,y)$ is determined by $x$ and $y$ (obviously), but also by the coefficients of $f_i$! When $f_i$ has very small coefficients, the number field sieve becomes asymptotically faster, from 

$$
\exp ( (1.923 + o(1))( (\log n)^{1/3}(\log \log n)^{2/3})
$$

to 

$$
\exp ( (1.526 + o(1))( (\log n)^{1/3}(\log \log n)^{2/3}).
$$

When one chooses a prime $p$ very close to $2^k$, it becomes easy to find a sparse polynomial that has a root modulo $p$. In your example, $p = 2^{2048} - 1942289$, we can find the degree $8$ polynomial 

$$
x^8 - 1942289,
$$

since $2^{2048} - 1942289 = (2^{256})^8 - 1942289$. This polynomial has very small coefficients, that render the number field sieve for the discrete logarithm in this field much faster that it would be for a random 2048-bit prime.