For any commutative ring $R,$ there is a unique homomorphism $f : \Bbb Z\to R,$ as specifying that $1\mapsto 1$ tells you where each element of $\Bbb Z$ must map to. Concretely, for any positive $n\in\Bbb Z,$ you have $n = 1 + \dots + 1$ ($n$ times,) so that $$f(n) = f(1 + \dots + 1) = f(1) + \dots + f(1) = n\cdot f(1).$$
You also know that $f(0) = 0,$ and if $n\in\Bbb Z$ is negative, then $f(n) = f(-(-n)) = -f(-n).$ Thus, there is exactly one morphism $f : \Bbb Z\to R,$ because the image of any element is determined by where $1$ is sent and the ring homomorphism rules.
Now, let's examine what it takes to define a morphism $\Bbb Z[x]\to R.$ We already know that we don't have a choice for where $\Bbb Z\subseteq\Bbb Z[x]$ is sent. What about $x$? Well, it turns out we can send $x$ to any element of $R$ that we want.
Suppose that $r\in R.$ If $f : \Bbb Z[x]\to R$ is a ring homomorphism sending $x$ to $r,$ then the ring homomorphism properties imply that we must have
\begin{align*}
f\left(\sum_{i = 0}^n a_i x^i\right) &= \sum_{i = 0}^nf\left( a_i x^i\right)\\
&= \sum_{i = 0}^n f(a_i) f(x^i)\\
&= \sum_{i = 0}^n f(a_i) f(x)^i\\
&= \sum_{i = 0}^n f(a_i) r^i.
\end{align*}
Every element of $\Bbb Z[x]$ is of the form $\sum_{i = 0}^n a_i x^i$ for some $n$ and some collection of integers $a_i,$ so we see that specifying where $x$ is sent determines the entire homomorphism. In particular, it is given by
$$
p(x) = \sum_{i = 0}^n a_i x^i \mapsto \sum_{i = 0}^n f(a_i) r^i = p(r).
$$
Conversely, setting $f(x) = r$ and extending in the above way is always a ring homomorphism. That is, let $r\in R$ and define
\begin{align*}
f : \Bbb Z[x]&\to R\\
\sum_{i = 0}^n a_i x^i &\mapsto \sum_{i = 0}^n g(a_i) r^i,
\end{align*}
where $g : \Bbb Z\to R$ is the unique ring homomorphism from paragraph one. We still have $f(1) = g(1) = 1$ and for any $n,m\in\Bbb Z\subseteq\Bbb Z[x],$ we have $$f(n + m) = g(n + m) = g(n) + g(m) = f(n) + f(m).$$ Suppose we have two arbitrary polynomials $\sum_{i = 0}^n a_i x^i,$ $\sum_{i = 0}^m b_i x^i.$ Then without loss of generality $m\leq n,$ and we can say that $\sum_{i = 0}^m b_i x^i = \sum_{i = 0}^n b_i x^i,$ where we define $b_j = 0$ for $j > m.$ Then we have
\begin{align*}
f\left(\sum_{i = 0}^n a_i x^i + \sum_{i = 0}^n b_i x^i\right) &= f\left(\sum_{i = 0}^n (a_i + b_i) x^i\right)\\
&= \sum_{i = 0}^n g(a_i + b_i)r^i\qquad\textrm{(by definition)}\\
&= \sum_{i = 0}^n \left(g(a_i) + g(b_i)\right)r^i\\
&= \sum_{i = 0}^n (g(a_i)r^i + g(b_i)r^i)\\
&= \sum_{i = 0}^n g(a_i)r^i + \sum_{i = 0}^n g(b_i)r^i\\
&= f\left(\sum_{i = 0}^n a_i x^i\right) + f\left(\sum_{i = 0}^n b_i x^i\right).
\end{align*}
You can check similarly that $$f\left(\left(\sum_{i = 0}^n a_i x^i\right)\cdot\left(\sum_{i = 0}^n b_i x^i\right)\right) = f\left(\sum_{i = 0}^n a_i x^i\right)\cdot f\left(\sum_{i = 0}^n b_i x^i\right),$$
so that the map defined is indeed a ring homomorphism.
The case of two variables is similar - a ring homomorphism is completely determined by where each variable is sent, and any choice of $r,s\in R$ gives a ring homomorphism with $x\mapsto r$ and $y\mapsto s.$
What's happening is that $\Bbb Z[x,y]$ is the free commutative ring on two generators $x$ and $y,$ which means essentially what I stated above - a ring homomorphism from $\Bbb Z[x,y]$ is given by a choice of where $x$ and $y$ will be sent, and any choices will work. To make sure that this defines a map on the quotient you want, you need to further specify that the images of $x$ and $y$ satisfy the given relation - i.e., because $x^3 + y^2 - 1 = 0$ in $\Bbb Z[x,y]/(x^3 + y^2 - 1),$ we must have $f(x)^3 + f(y)^2 - 1 = 0$ as well.
Best Answer
No, not at all. A homomorphism must take nilpotent elements to zero, since $\Bbb Z$ has no proper nilpotents. The matrix that’s all zero except for a $1$ in the upper right corner must thus be taken to $0$. Similarly for the matrix with $1$ in the lower left. But their sum squares to the identity matrix, so your homomorphism is zero. (There are much better abstract proofs.)