= Solution
For an integer $m>1$, let $\alpha_m$ be a root of
$$
f_m(X)=X^2-2mX+m.
$$
Its roots are
$$
R_m=m+\sqrt{m^2-m}>1,
\qquad
r_m=m-\sqrt{m^2-m}\in(0,1).
$$
The polynomial is <irreducible polynomial>[irreducible] over $\mathbb Q$: its discriminant is $4m(m-1)$, and the product of the coprime consecutive integers $m$ and $m-1$ cannot be a square unless both are squares, which is impossible for consecutive positive squares beyond $0,1$.
The <height-Mahler measure formula> gives
$$
H(\alpha_m)^2=R_m.
$$
Both <algebraic conjugates> of $\alpha_m+1$ exceed $1$, so
$$
H(\alpha_m+1)^2
=(R_m+1)(r_m+1)
=f_m(-1)=3m+1.
$$
Consequently
$$
\frac{H(\alpha_m+1)}{H(\alpha_m)}
=\sqrt{\frac{3m+1}{R_m}}
\longrightarrow\sqrt{\frac32}>1.
$$
Fix, for example, $\delta=1/10$. For all sufficiently large $m$, the ratio is larger than $1+\delta$ by a fixed margin, while $H(\alpha_m)\to\infty$. Hence, for every $C>0$, some sufficiently large $m$ satisfies
$$
H(\alpha_m+1)>(1+\delta)H(\alpha_m)+C.
$$
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