Let be $\phi$ a scalar field which depends for simplicity only on coordinates $x^\mu$ (i.e $\phi = \phi(x^\mu)$).
Let's do an infinitesimal translation:
$x'^\mu = x^\mu + \delta x^\mu$
Thus, the field changes, by definition, as: $\delta\phi =\dfrac{\partial\phi}{\partial x^\mu}\delta x^\mu \equiv \delta x^\mu \partial_{\mu}\phi \tag{a}\label{eq:a} $.
But, on the other hand, we already now that, for a scalar field, we have:
$\phi'(x') = \phi(x) $. If we expand this as a Taylor series we will have:
$\phi'(x+\delta x) = \phi'(x)+\delta x^\mu \partial_\mu \phi(x) = \phi(x)$ where we've used that, at first order, $\partial_\mu\phi'(x)\approx \partial_\mu\phi(x)$.
But then we conclude that $\delta\phi = \phi'(x)-\phi(x)=-\delta x^\mu \partial_\mu \phi(x) \tag{b}\label{eq:b}$
\eqref{eq:a} is not equal to \eqref{eq:b}. Both differs in a sign. What is wrong? What is the right way for finding the variation?
When we derive the Euler-Lagrange equations from the variation of the action we make use of \eqref{eq:a}, so I figure that this is the correct one.
There is no contradiction. You are comparing active and passive transformations of the field. For an extremely basic example, consider $x\in\mathbb R$ and $\phi(x)=x$. Then $x'=x+\epsilon$ corresponds exactly to $\phi'(x)=x-\epsilon$. The opposite signs of the derivate occurs since the active and passive transformations are inverse to one another.
Since the Euler-Lagrange equations are stationary equations (i.e. setting functional derivative to zero) it actually does not matter much which convention one is using, since $-0=0$. That being said, the derivation used on wikipedia is to consider $\phi(x+\delta x)$ (in your notation) instead of $\phi'(x)$.