Interpret the equation correctly.

The given condition is $nA^n + mB^m = I$ for $n,m \in \{1,2,\dots,10\}$.

We need to count the number of ordered pairs $(n,m)$ satisfying this.

Compute powers of $A$.

Given $A=\begin{pmatrix}2&-2\\1&-1\end{pmatrix}$

First compute $A^2$: $A^2=\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\begin{pmatrix}2&-2\\1&-1\end{pmatrix}$ $=\begin{pmatrix}4-2&-4+2\\2-1&-2+1\end{pmatrix}$ $=\begin{pmatrix}2&-2\\1&-1\end{pmatrix}=A$

So $A$ is idempotent, hence $A^n=A \quad \text{for all } n\ge 1.$

Therefore

Compute powers of $B$.

Given $B=\begin{pmatrix}-1&2\\-1&2\end{pmatrix}$

First compute $B^2$: $B^2=\begin{pmatrix}-1&2\\-1&2\end{pmatrix}\begin{pmatrix}-1&2\\-1&2\end{pmatrix}$ $=\begin{pmatrix}1-2&-2+4\\1-2&-2+4\end{pmatrix}$ $=\begin{pmatrix}-1&2\\-1&2\end{pmatrix}=B$

So $B$ is also idempotent, hence $B^m=B \quad \text{for all } m\ge 1.$

Therefore

Form the equation $nA^n+mB^m=I$.

Adding the two matrices, nA^n+mB^m=egin{pmatrix}2n-m&-2n+2m\\n-m&-n+2m\end{pmatrix}.

This must equal the identity matrix $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}.$

Hence we get the system: $2n-m=1 \quad ...(1)$ $-2n+2m=0 \quad ...(2)$ $n-m=0 \quad ...(3)$ $-n+2m=1 \quad ...(4)$

Solve the system.

From (3), $n=m.$

Substitute into (1): $2n-n=1 \Rightarrow n=1.$ Therefore $m=1.$

Check in (4): $-1+2=1,$ true.

So the only ordered pair is $(n,m)=(1,1).$

Count the number of such pairs.

There is exactly one element in the set.

Therefore, the required number is $\boxed{1}.$