17.79 (2010)

Yes, moreover, there is a continuum of such groups (V. S. Atabekyan, Infinite simple groups satisfying an identity, Dep. VINITI no. 5381-V86, Moscow, 1986 (Russian)). Another example was suggested by D. Osin in a letter of 31 August 2013: a free group $G$ in the variety $\mathfrak{M}$ defined by the law $x^n y = yx^n$ is a central extension of a free Burnside group of exponent $n$ such that the centre of $G$ is a free abelian group of countable rank (I. S. Ashmanov, A. Yu. Olshanskii, Izv. Vyssh. Uchebn. Zaved. Mat., 1985, no. 11 (1985), 48–60 (Russian)). Let $x_1, x_2, \dots$ be a basis in $Z(G)$. By adding to $G$ the relations $x_i^{r_i} = 1$ for all $i$ we obtain a periodic group of unbounded exponent generating a proper variety (contained in $\mathfrak{M}$).