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Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere.

Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere, but here is a proof that at least it has the cohomology with coefficients in a field $k$ of a $2$-sphere, and is therefore not contractible.

I think it is a standard fact that the cohomology $H^n(B\mathcal{C},k)$ of the classifying space of $\mathcal{C}$ is equal to the extension group $\text{Ext}^n(\mathbf{k},\mathbf{k})$ in the category of functors from $\mathcal{C}$ to $k$-vector spaces, where $\mathbf{k}$ is the constant functor taking the value $k$.

For each object $V$ of $\mathcal{C}$, there is a projective functor $P_V$ whose value on an object $U$ is the vector space with basis $\mathcal{C}(V,U)$, and a morphism $\alpha:V\to V'$ induces a morphism of functors $\alpha^\ast:P_{V'}\to P_V$ by composition.

A straightforward calculation shows that $$0\longrightarrow P_X\oplus P_X\stackrel{\pmatrix{\beta^\ast_1\\\beta^\ast_2}}{\longrightarrow} P_Y\stackrel{\alpha^\ast_1-\alpha^\ast_2}{\longrightarrow} P_X\longrightarrow\mathbf{k}\longrightarrow0$$ is a projective resolution of the constant functor, and applying the functor $\text{Hom}(-\mathbf{k})$ to the projective terms to calculate $\text{Ext}^*(\mathbf{k},\mathbf{k})$ gives $$k\stackrel{0}{\longrightarrow}k\stackrel{\pmatrix{1&1}}{\longrightarrow}k^2\longrightarrow0,$$ so $\text{Ext}^*(\mathbf{k},\mathbf{k})$ is one-dimensional in degrees zero and two, and zero in all other degrees.

I'll leave my original example here for a while, but here's a much simpler one.

Let $\mathcal{C}$ be the category with three objects $W$, $X$ and $Y$, with the only non-identity maps being $\alpha_1,\alpha_2:X\to Y$, $\beta:W\to X$, and the compositions $\alpha_i\beta$.

Then $X$ satisfies conditions (1) and (2), and since there are no infinite composable chains of non-identity morphisms, it really is easy to calculate that the nerve is homotopy equivalent to a circle. Or alternatively, the full subcategory containing $X$ and $Y$ is a retract, and its nerve is a circle, so the nerve of $\mathcal{C}$ can't be contractible.

Original example: Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere.

Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere.

I'll leave my original example here for a while, but here's a much simpler one.

Let $\mathcal{C}$ be the category with three objects $W$, $X$ and $Y$, with the only non-identity maps being $\alpha_1,\alpha_2:X\to Y$, $\beta_1,\beta_2:W\to X$, and the four compositions $\alpha_i\beta_j$.

Then $X$ satisfies conditions (1) and (2), and since there are no infinite composable chains of non-identity morphisms, it really is easy to calculate that the nerve is homotopy equivalent to a figure of eight. Or alternatively, the full subcategory containing $X$ and $Y$ is a retract, and its nerve is (homotopy equivalent to) a circle, so the nerve of $\mathcal{C}$ can't be contractible.

Original example: Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere.

I'll leave my original example here for a while, but here's a much simpler one.

Let $\mathcal{C}$ be the category with three objects $W$, $X$ and $Y$, with the only non-identity maps being $\alpha_1,\alpha_2:X\to Y$, $\beta:W\to X$, and the compositions $\alpha_i\beta$.

Then $X$ satisfies conditions (1) and (2), and since there are no infinite composable chains of non-identity morphisms, it really is easy to calculate that the nerve is homotopy equivalent to a circle. Or alternatively, the full subcategory containing $X$ and $Y$ is a retract, and its nerve is a circle, so the nerve of $\mathcal{C}$ can't be contractible.

Original example: Let $\mathcal{C}$ be the category with two objects $X$ and $Y$, generated by morphisms $\alpha_1,\alpha_2:X\to Y$ and $\beta_1,\beta_2:Y\to X$ subject to relations $\beta_i\alpha_j=\text{id}_X$ for all $i,j$.

So the only non-identity morphisms other than $\alpha_1,\alpha_2,\beta_1,\beta_2$ are the compositions $\alpha_i\beta_j:Y\to Y$.

According to my calculations, the nerve is homotopy equivalent to a $2$-sphere.