It is possible to use just 1 variable "$L(x)$=$x$ has 1 favourite toy", or even two variables "$L(x,y)$=$x$ has 1 favourite $y$" however it is a better learning experience to specify what you mean by exactly 1.
A sufficient way to express "There is a dog who has exactly one favorite toy," in symbolic logic is the following $(1)$.
\begin{equation}\tag{1}
\exists x\big(\exists y(D(x)\land T(y)\land F(x,y)\land\forall z((F(x,z)\land T(z))\rightarrow z=y))\big),
\end{equation}
where $D(x)$ expresses "$x$ is a dog", $T(y)$ "$y$ is a toy", and $F(x,y)$ "$x$'s favorite toy is $y$."
The way $(1)$ works is that it asserts that there is a dog $x$, a toy $y$ and the dogs favorite toy is $y$. Then it goes on to assert, that any toy which is the dogs favorite, is the same toy as the first. Thus there is only 1 toy.
For your future reference, I have included the definitions of less than, greater than or equal to, and exactly 1 below this answer.
Less than n:
\begin{equation}\tag{2}
\forall x_1\dots\forall x_{n-1}\forall x_n((x_1\neq x_2\land\dots\land x_{n-1}\neq x_n)\rightarrow \lnot P(x_1)\lor\dots\lor\lnot P(x_{n-1})\lor\lnot P(x_n))
\end{equation}
Greater than or equal to n:
\begin{equation}\tag{3}
\exists x_1\dots\exists x_{n-1}\exists x_n(x_1\neq x_2\land\dots\land x_{n-1}\neq x_n\land P(x_1)\land\dots\land P(x_{n-1})\land P(x_n))
\end{equation}
Exactly 1
\begin{equation}\tag{4}
\exists x(P(x)\land\forall y(P(y)\rightarrow y=x))
\end{equation}
